Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding the number of real roots of an unusual(!) equation How many real roots does the below equation have?
\begin{equation*}
\frac{x^{2000}}{2001}+2\sqrt{3}x^2-2\sqrt{5}x+\sqrt{3}=0
\end{equation*}
A) 0 B) 11 C) 12 D) 1 E) None of these
I could not come up with anything.
(Turkish Math Olympiads 2001)
| We have that $x^{2000} \geq 0$, because squares are nonnegative.
Further, we have $(x-\frac{1}{2}\sqrt{\frac{5}{3}})^2 \geq 0$. This gives $x^2-\sqrt{\frac{5}{3}}x+\frac{5}{12} \geq 0$
Therefore $2 \sqrt{3}x^2-2\sqrt{5}x+\frac{10}{12}\sqrt{3} \geq 0$
Therefore $2 \sqrt{3}x^2-2\sqrt{5}x+\sqrt{3} > 0$
Therefore $\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Determinant properties Prove without expanding:
\begin{equation}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3 & c^3\end{vmatrix} = (ab + ac + bc)(b - a)(c - a)(c - b)\end{equation}
*I tried to zero some elements and expand until I reach the Right hand side.
*Also tried C1-C3, C2-C3 then decompose the determinant into t... | I found a long proof but that's the only way I could answer it.
\begin{gather*}
\begin{bmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{bmatrix} \\
= \begin{bmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3 - a^3\end{bmatrix} \\
= (b-a)(c-a)\begin{bmatrix}1&0&0\\a^2&b+a&c+a\\a^3&b^2+ab+a^2&c^2+ac+a^2\end{bmatrix} \\
= (b-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the smallest integer that is divisible by exactly $X$ perfect squares. Is there a method to find the smallest integer divisible by exactly $X$ perfect squares?
Example: find the smallest positive integer divisible by exactly 2015 perfect squares.
I've been trying to figure this out, but I haven't made much progres... | First of all, there won't be a simple formula.
If $X$ is prime, the smallest number is $2^{2X-2}$. So the smallest number divisible by example five perfect squares is $2^{8}=256$.
More generally, it will depend on factorizations of $X$.
In general, if $N$ is the smallest integer with exactly $X$ positive integer divis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can $T:P_3\rightarrow P_4$ be a function of two variables? I'm given the following question:
Denote by $P_n$ the set of polynomials in the variable $x$ of degree at most $n$. The usual basis for $P_n$ is given by $\left\{1,x,x^2,\ldots,x^n\right\}$.
Define the linear transformation $T:P_3\rightarrow P_4$ by $T(p(x))=3... | We can re-write the formula for $T$ as
\begin{align*}
T(a+bx+cx^2+dx^3)
&= \frac{d}{dx}(a+bx+cx^2+dx^3)+\int_0^x(a+bt+ct^2+dt^3)\,dt \\
&= b+2\,cx+3\,dx^2+\left[at+\frac{b}{2}t^2+\frac{c}{3}t^3+\frac{d}{4}t^4\right]_0^x \\
&= b+2\,cx+3\,dx^2+ax+\frac{b}{2}x^2+\frac{c}{3}x^3+\frac{d}{4}x^4-0 \\
&= b+(2\,c+a)x+\left(3\,d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving equations including floor function. I got a little trouble solving equations that involve floor function in an efficient way.
For example :
$$
\left\lfloor\frac{x+3}{2}\right\rfloor = \frac{4x+5}{3}
$$
In the one above, I get that you basically let $$ \frac{4x+5}{3} = k $$ and then inserting $k$ in the left sid... | Since $\dfrac{4x+5}3 = \left\lfloor \dfrac{x+3}2 \right \rfloor$ is an integer, we need $3$ to divide $4x+5$, i.e.,
$$\dfrac{4x+5}3 = m \in \mathbb{Z} \implies x = \dfrac{3m-5}4 \text{ where }m \in \mathbb{Z}$$
Hence,
$$\left\lfloor \dfrac{x+3}2 \right\rfloor = m \implies \dfrac{x+3}2 = m + e \implies x+3 = 2m+2e \impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find all the right inverses of a matrix How do I find the right inverse of a non square matrix?
The matrix i have is
$$M =
\begin{bmatrix}
1 & 1 & 0 \\
2 & 3 & 1\\
\end{bmatrix}$$
Im really not sure how to even start this?
| Right inverse means a matrix $A_{3 \times 2}$ such that $MA=I_{2 \times 2}$. So you are looking for a matrix $A=\begin{pmatrix}x&p\\y&q\\z&r\end{pmatrix}$ such that
$$MA =
\begin{pmatrix}
1 & 1 & 0 \\
2 & 3 & 1\\
\end{pmatrix}\begin{pmatrix}x&p\\y&q\\z&r\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$
This gives t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1276049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $. I have the equation $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $.
I tried to square both sides, but then I got a more difficult equation:
$$
2 x - 11 + 2 \sqrt{x^{2} - 11 x - 28} = 1.
$$
Can someone tell me what I should do next?
| squaring gives
$$x-4+x-7+2\sqrt{x-4}\sqrt{x-7}=1$$
$$\sqrt{x-4}\sqrt{x-7}=12-2x$$
$$(x-4)(x-7)=36+x^2-12x$$
$$x=8$$
but $8$ fulfills not our equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find the remainder when $3^{89}7^{86}$ is divided by $17$
Find the remainder when $3^{89}7^{86}$ is divided by $17$.
I guess the problem is to be solved by congruencies. But unfortunately, I have no clear conception about it.
Can someone please explain it.
Thank you.
| You can check $3$ has order $16$ modulo $17$. This means $3^{16}\equiv 1\mod 17$, but $3^k\not\equiv 1$ for $1\le k<16$ Actually, Little Fermat ensures $a^{16}\equiv 1 \mod 16$ for any $a$ not divisible by $17$, and Lagrange's theorem says the order of $a$ is a divisor of $16$. Also $3^8\equiv -1$.Thus:
$$3^{89}=\bigl(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving complex numbers equation $z^3 = \overline{z} $ We have the following equation:
$$z^3 = \overline{z} $$
I set z to be $z = a + ib$ and since I know that $ \overline{z} = a - ib$. I was trying to solve it by opening the left side of the equation.
$$ z^3 = (a+ib)^3 \Rightarrow $$
$$ [a^2+b^2+i(ab + ba)](a+ib) \Rig... | You have a good start. Rewrite equation as $z^3-\bar{z} =0$, now do $z=a+bi$, so we get $$a^3-3b^2a-a+i(3a^2b-b^3-b) = 0$$
Now both the imaginary and the real part must be equal to zero, so we get the following system of equations $$a^3-3b^2a-a=0 \wedge 3a^2b-b^3-b=0$$
Factoring gives:
$$a(a^2-3b^2-1)=0 \wedge b(3a^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
} |
What is the biggest $n$ in $4^n$ that divides $7^{2048} - 1$? A few days ago I stumbled on the following question, it was used in the Museum of mathematics masters tournament:
What is the biggest integer $n$ in $4^n$, that divides $7^{2048} - 1$?
a) 1
b) 3
c) 5
d) 7
It was not phrased exactly like this, but it is simil... | You can write
\begin{align*}
7^{2048} - 1 &= 7^{2048} - 1^{2048} \\
&= (7^{1024} + 1)(7^{512} + 1)(7^{256} + 1)(7^{128} + 1)(7^{64} + 1)(7^{32} + 1)(7^{16} + 1)(7^{8} + 1)(7^{4} + 1)(7^{2} + 1)(7 + 1)(7-1)
\end{align*}
For all $k \geq 2$, $7^{2^k} + 1 \equiv (-1)^{2^k} + 1 \equiv 2 \mod 4$, so each term of the form $(7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Solve the equation $\log_{2} x \log_{3} x = \log_{4} x$ Question:
Solve the equations
a)
$$\log_{2} x + \log_{3} x = \log_{4} x$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
Attempted solution:
The general idea I have been working on is to make them into the same base, combine them and take the inverse to get the soluti... | rewrite the second equation as $$\frac{\ln(x)}{\ln(2)}\cdot\frac{\ln(x)}{\ln(3)}= \frac{\ln(x)}{2\ln(2)}$$
and the first equation as
$$\frac{\ln(x)}{\ln(2)}+\frac{\ln(x)}{\ln(3)}=\frac{\ln(x)}{2\ln(2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Listing all the possible echelon forms of a $3\times 3$ matrix My approach.
Let $a$ denote a leading entry and $b$ be any value.
The possible echelon forms of a $3\times 3$ matrix are:
$$\begin{bmatrix} a & b & b \\ 0 & a & b \\ 0 & 0 & a\end{bmatrix}, \begin{bmatrix} a & b & b \\ 0 & a & b \\ 0 & 0 & 0\end{bmatrix}, \... | here are the rrefs of $3 \times 3:$
of rank $3:$ $\quad \pmatrix{1&0&0\\0&1&0\\0&0&1}$
of rank $2: \quad\pmatrix{1&0&x\\0&1&x\\0&0&0} , \pmatrix{1&x&0\\0&0&1\\0&0&0}, \pmatrix{0&1&0\\0&0&1\\0&0&0}$
of rank $1: \quad\pmatrix{1&x&x\\0&0&0\\0&0&0}, \pmatrix{0&1&x\\0&0&0\\0&0&0}, \pmatrix{0&0&1\\0&0&0\\0&0&0}$
of rank $0:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the integer x: $x \equiv 8^{38} \pmod {210}$ Find the integer x: $x \equiv 8^{38} \pmod {210}$
I broke the top into prime mods:
$$x \equiv 8^{38} \pmod 3$$
$$x \equiv 8^{38} \pmod {70}$$
But $x \equiv 8^{38} \pmod {70}$ can be broken up more:
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod {10}$$
But $x \equiv... | we have $$64\equiv 8^{38}\mod 210$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 7
} |
$ \lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15} $ Can someone show me how is possible to prove that
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15}
\end{equation*}
but without Taylor series. One can use L'Hospital rule if necessary. I was not abl... | @Paramanand. Thank you very much. Your method is powerfull. After reading
it, I am able now to combine it with some unpublished computations I did.
My favorite method is to do not use LHR nor Taylor expansion whenever
possible and to came back all the computations to the basic limits only. For
example, to compute the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the limit of: $\lim_{x\to 1}\left(\frac{\sqrt{2x-1}-1}{x^2-1}\right)$
Find the limit of: $\lim\limits_{x\to 1}\left(\frac{\sqrt{2x-1}-1}{x^2-1}\right)$
I tried multiplying by the conjugate of $\sqrt{2x-1}-1$ ,but I obtain
$\frac{2x-2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{2x-1}+1\right)}$,
which is again z... | Another approach:
With the following substitution: $$\begin{aligned}\sqrt{2x-1}=t\implies x=&\ \frac{t^2+1}2\\\\\frac1{x+1}=&\ \frac2{t^2+3}\\\\\frac1{x-1}=&\ \frac2{t^2-1}=\frac2{(t-1)(t+1)}\ ,\end{aligned}$$
we have:
$$\lim_{x\to 1}\frac{\sqrt{2x-1}-1}{x^2-1}=\lim_{x\to 1}\frac{\sqrt{2x-1}-1}{(x-1)(x+1)}=\lim_{t\to 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Probability of $ax^2 + bx + c = 0$ having real solutions
$a$, $b$, $c$ are random integer numbers between $1$ and $100$ (including $1$ and $100$, and uniformly distributed). What is the probability that the equation $ax^2 + bx + c = 0$ has real solutions?
This is from a final high school math exam, and I don't know h... | Interesting question. What kind of class was this?
We can approximate this in the continuous case by letting $(a, b, c)$ reside in the open unit cube in the positive octant. Given any value of $b$ in the interval $(0, 1)$, the allowable values of $(a, c)$ are those in the region of the first-quadrant unit square "ins... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 0
} |
How do you find the $4\times 4$ matrix corresponding to the transformation T with respect to the basis? If the transformation $T$ acting on the vector space $A \in Mat_{2,2}$ is given by $T(A)=CA$, where
$
C= \left( \begin{array}{cc}
1 & 2 \\
3 & 4 \end{array} \right)
$ how would you find the $4\times 4$ matrix corre... | Hint:
If we take the standard basis $\Biggl\{\begin{pmatrix}1&0\\0&0\end{pmatrix}, \begin{pmatrix}0&1\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\1&0\end{pmatrix}, \begin{pmatrix}0&0\\0&1\end{pmatrix}\Biggr\}$ for $M_{2,2}$,
then $T(v_1)=\begin{pmatrix}1&0\\3&0\end{pmatrix}, T(v_2)=\begin{pmatrix}0&1\\0&3\end{pmatrix}, T(v_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $A$ be a complex $2$ by $2$ matrix having distinct eigenvalues $a, b$. Show that $A^n =\frac{ a^n}{a - b}(A - bI) + \frac{b^n}{b - a}(A - aI)$.
Let $A\in\mathscr{M}_{2\times 2}(\mathbb{C})$ be a matrix having distinct eigenvalues $a\neq b$. Show that, for all $n > 0$,
\begin{equation*}
A^n =\frac{ a^n}{a - b}(A ... | We have
$$A = X \begin{bmatrix}a & 0\\0 & b\end{bmatrix} X^{-1}$$
This means
$$A^n = X \begin{bmatrix}a^n & 0\\0 & b^n\end{bmatrix} X^{-1}$$
We have
$$A-bI = X \begin{bmatrix}a-b & 0\\0 & 0\end{bmatrix} X^{-1} \text{ and }A-aI = X \begin{bmatrix}0 & 0\\0 & b-a\end{bmatrix} X^{-1}$$
Hence,
$$\dfrac{a^n}{a-b}\left(A-bI\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the largest triangle inscribed in the unit circle Among all triangles inscribed in the unit circle, how can the one with the largest area be found?
| Given any triangle $\triangle ABC$ of sides $a,b$ and $c$, let $R$ be its circumradius and $\mathcal{A}$ be its area. We have this interesting identity:
$$4 R \mathcal{A} = abc$$
When $ABC$ is inscribed inside the unit circle, $R = 1$ and by $GM \le AM$, we have
$$\mathcal{A} = \frac14 abc \le \frac14 \left(\frac{a^2+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 8,
"answer_id": 0
} |
What will be the equation of side $BC$.
The equation of two equal sides $AB$ and $AC$ of an isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$ respectively . What will be the equation of the side $BC$ if the area of the triangle $\triangle ABC$ is $5$ square units.
$a.)x+3y-1=0\\
b.)x-3y+1=0\\
c.)2x-y-5=0\\
\color{gre... | Let the lines: $x+y=5$ & $7x-y=3$ represent AC & AB. Then the acute angle $\theta$ between the equal sides AB & AC is given as $$\tan \theta=\left|\frac{7-(-1)}{1+7(-1)}\right|=\left|\frac{8}{-6}\right|=\frac{4}{3} \implies \sin \theta=\frac{4}{5}$$ The length of two sides AB & AC of isosceles triangle are equal hence,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove that $f=x^6+ax+5$ is reducible over $\mathbb{Z_7},\forall a\in\mathbb{Z_7}$ We have $f=x^6+ax+5\in\mathbb{Z_7}$ and we have to show that it is reducible on $\mathbb{Z_7}$, $\forall a\in\mathbb{Z_7}$. Here are all my steps:
For $a=0$ we'll get $f=x^6+5\in\mathbb{Z_7}$. But $\forall x\in\mathbb{Z_7}$ with $x\neq0\... | Note that $f(a^{-1})=0$ as you noted before when $a$ is not $0$. Then we know that $(x-a^{-1})$ is a factor. We can do a "complete the sextic" approach on the function as follows:
$f(x) = (x-a^{-1})(x^5) = x^6-a^{-1}x^5$
$f(x) = (x-a^{-1})(x^5+a^{-1}x^4) = x^6 + a^{-2}x^4$
Verify that $f(x) = (x-a^{-1})(x^5+a^{-1}x^4+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
First order differential equation: did i solve this equation right So i'm trying to solve:
$$x^2\frac{dy}{dx} + 2xy = y^3$$
I'm given this differential equation, that Bernoulli equation:
$$\frac{dy}{dx} + p(x)y = q(x)y^{n} $$
I think i've solved it and got
$$ u = \frac{2}{5x} +Cx^4$$
I'm just not sure i am right i... | $$
x^2y' + 2xy = y^3
$$
first thing to notice is
$$
\dfrac{d}{dx}x^2y = x^2y' + 2xy
$$
so we have
$$
\dfrac{d}{dx}x^2y = y^3
$$
let $v = x^2y$
we then have
$$
\dfrac{dv}{dx} = \left(\frac{v}{x^2}\right)^3 = \frac{1}{x^6}v^3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
why $ \sin \theta = \frac{7}{8} \cos \theta$? I have an example:
$$ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{7}{8} $$
and then this equation is true? why there is cos multiplied?:
$$ \sin \theta = \frac{7}{8} \cos \theta$$
| Your initial equation is $$\frac ab = \frac 78.$$ If you multiply this equation by $b$, you get
$$\frac ab \cdot b = \frac 78\cdot b\\
\frac{a\cdot b}{b} = \frac78\cdot b\\
a =\frac78\cdot b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Difficult inverse tangent identity
Prove that:
$$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right) = \frac{\pi}{4} - \frac{1}{2}\arccos(x), -\frac{1}{\sqrt{2}} \le x \le 1$$
I'd multiply the inside of $\arctan$ by the conjugate of the denominator.
I get:
$$\arctan\left(\frac{1 - 1\sqrt{... | Let $\dfrac12\arccos x=y\implies x=\cos2y$
and $-\dfrac1{\sqrt2}\le x\le1\implies-\dfrac1{\sqrt2}\le\cos2y\le1$
Using the definition of Principal values of $\arccos,0\le2y\le\dfrac{3\pi}4\iff0\le y\le\dfrac{3\pi}8 \ \ \ \ (1)$
$\implies\sin y,\cos y\ge0$
$\implies\sqrt{1-x}=\sqrt{1-\cos2y}=+\sqrt2\sin y$
$\implies\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How do you solve this quadratic equation? The number of values of a for which
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
is an identity in x is?
Here's how much I was able to solve through:-
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
$$ ((a-1)(a-2))x^2 + ((a-3)(a-2))x + (a+2)(a-2) = 0$$
$$ (a-2)[(a-1)x^... | Since
$$(a-1)x^2+(a-3)x+a+2$$
is a quadratic polynomial, it has, at most, two roots, then $$(a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
is an identity in $x$ only if $a=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim_{n\to\infty}nI_n$ with $I_n=\int_0^1\frac{x^n}{x^2+3x+2}dx$ We have to evaluate: $$\lim_{n\to\infty}nI_n$$ with $$I_n=\int_0^1\frac{x^n}{x^2+3x+2}\:dx.$$
There is an elegant way to solve this problem?
Here is all my steps:
*
*My first ideea was to find a recurrence relation such that:
$$I_{n+2}+3... | We may just integrate by parts,
$$
\begin{align}
I_n=\int_0^1\frac{x^n}{(x+1)(x+2)}dx&=\left. \frac{x^{n+1}}{(n+1)}\frac{1}{(x+1)(x+2)}\right|_0^1+\frac{1}{(n+1)}\int_0^1\frac{(2x+3)\:x^{n+1}}{(x+1)^2(x+2)^2}\:dx\\\\
&=\frac1{6(n+1)}+\frac{1}{n+1}\int_0^1\frac{(2x+3)}{(x+1)^2(x+2)^2}\:x^{n+1}dx\\\\
&=\frac1{6(n+1)}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Why cannot $(2x^2 + x)^2$ be simplified to $2x^4 + x^2$? So, I want to simplify an equation : $(2x^2 + x)^2$.
I thought this would simplify to $2x^4 + x^2$
But, if you input a value for $x$, the answers do not equal. For example, if you input $x = 3$, then:
$$(2x^2+x)^2
= 21^2
= 441$$
AND:
$$2x^4 + x^2
= 2(82) + 9
= 1... | In general, $(a + b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2$. So, in your case,
$$(2x^2 + x)^2 = (2x^2)^2 + 2(2x^2\cdot x) + x^2 = 4x^4 + 4x^3 + x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Why does $e^{-(x^2/2)} \approx \cos[\frac{x}{\sqrt{n}}]^n$ hold for large $n$? Why does this hold:
$$
e^{-x^2/2} = \lim_{n \to \infty} \cos^n \left( \frac{x}{\sqrt{n}} \right)
$$
I am not sure how to solve this using the limit theorem.
| For sure, this is not as elegant as previous answers but I love too much Taylor expansions !
Starting with $$ \cos \left( \frac{x}{\sqrt{n}} \right)=1-\frac{x^2}{2 n}+\frac{x^4}{24 n^2}-\frac{x^6}{720 n^3}+\frac{x^8}{40320
n^4}-\frac{x^{10}}{3628800 n^5}+O\left(x^{12}\right)$$ and raising to power $n$ (using binomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 6
} |
simple 2 sides inequality
$$2<\frac{x}{x-1}\leq 3$$
Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$
| you can rewrite $$2 < \frac x{x-1} = 1 + \frac 1{x-1} \le 3 \to 1 < \frac1 {x-1} \le 2 \tag 1$$ from the graph of $y = \frac1{x-1},$ we see that there are no solutions $(-\infty, 1)$ and $x - 1 > 0$ is necessary so $(1)$ is equivalent to $$\frac 12 \le x - 1< 1\to \frac32 \le x < 2. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
What is wrong with this integral reasoning? $$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx$$
We start by multiplying by $1=\frac{x}{x}$.
$$\int\frac{x^{2}+1}{x^{2}\sqrt{x^{4}+1}}xdx$$
Next, we use the substitution $u=x^{2}$;$\frac{du}{2}=xdx$.
$$\frac{1}{2}\int\frac{u+1}{u\sqrt{u^{2}+1}}du=\frac{1}{2}\int\frac{1}{\sqrt{u^{2}+... | The two expressions are equal up to a constant.
Take from your answer $$\frac{x^2(x^2+\sqrt{x^4+1})}{1+\sqrt{x^4+1}}$$
and multiply top and bottom by $\sqrt{x^4+1}-1$
you get
$$\frac{x^4-x^2+1+(x^2-1)\sqrt{x^4+1}}{x^2}=\frac{1}{2}\frac{(x^2-1+\sqrt{x^4-1})^2}{x^2}$$
So $$\frac{1}{2} \ln \frac{x^2(x^2+\sqrt{x^4+1})}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
$(1+i)^6$ in polar form $re^{i\theta}$ I used De Moivre's formula and got
$$
\left(\frac{\sqrt{2}}{2}\right)^6 \times
\cos\left(6 \times \frac{1}{4\pi}\right) +
i\sin\left(6 \times \frac{1}{4\pi}\right) =
\frac{1}{8} e^{\frac{3}{2\pi}}.
$$
But the answer is $8e^{\frac{3}{2\pi}}$, can someone explain where t... | $1+i = \sqrt{2}\left(\cos (\frac{\pi}{4}) + i\sin (\frac{\pi}{4})\right)= \sqrt{2}e^{i\frac{\pi}{4}} \Rightarrow (1+i)^6 = (\sqrt{2})^6\cdot e^{i(3\frac{\pi}{2})}=-8i$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How to do power series expansion What is the coefficient of $x^{11}$ in the power series expansion of $\frac 1{1-x-x^4}$?
How do I do power series expansions?
| suppose $$\frac{1}{1-x-x^4} = 1 + x + a_2x^2 + a_3x^3 + \cdots. $$ then we have $$1 =(1-x-x^4)(1 + x + a_2x^2 + a_3x^3 +\cdots)=1+(a_2-1)x^2 + (a_3-a_2)x^3+\\(a_4-a_3-1)x^4+\cdots+(a_n-a_{n-1} -a_{n-4})x^n+\cdots $$ equating the coefficients of $x^2, x^3, \cdots$, we find that
$$ a_0 = 1, a_1 = 1, a_2 = 1, a_3 =1, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Where am I wrong in $\int_0^1 x(1-x)^ndx$ We have $\int_0^1 x(1-x)^ndx=-\frac{x(1-x)^{n+1}}{n+1}|^{^1}_{_0}+\frac{(1-x)^{n+2}}{(n+1)(n+2)}|^{^1}_{_0}=-\frac{1}{(n+1)(n+2)}$
If I use substitution $u=1-x\Rightarrow \int_0^1 (1-u)u^n du=\frac{u^{n+1}}{n+1}|^{^1}_{_0}-\frac{u^{n+2}}{n+2}|^{^1}_{_0}=\frac{1}{(n+1)(n+2)}$
... | REVISED BY REQUEST
We begin with the integral $\int_0^1 x(1-x)^ndx$. Integrating by parts gives
$$\begin{align}
\int_0^1 x(1-x)^ndx&=\left.-\frac{x(1-x)^{n+1}}{n+1}\right|^{^1}_{_0}+\frac{1}{n+1}\int_0^1 (1-x)^{n+1}dx\\\\
&=\frac{1}{n+1}\int_0^1 (1-x)^{n+1}dx
\end{align}$$
Now, integrating the remaining term reveals t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\lim\limits_{n\to\infty}\prod\limits_{k=2}^{n}\frac{k^2+k-2}{k^2+k}$ I can't find the product of a sequence.
We have
$$\frac{(2+2)(2-1)}{2(2+1)}\frac{(3+2)(3-1)}{3(3+1)}...\frac{(k+2)(k-1)}{k(k+1)}$$
I am stuck with $$P=\frac{2(n+2)}{n^2(n-1)}$$ but that isn't correct. Can the squeeze theorem be used?
| HINT:
We have
$$\begin{align}
\prod_{k=2}^n\frac{k^2+k-2}{k^2+k}&=\prod_{k=2}^{n} \left(\frac{k+2}{k+1}\frac{k-1}{k}\right)\\\\
&=\prod_{k=2}^{n} \left(\frac{k+2}{k+1}\right)\prod_{k=2}^{n} \left(\frac{k-1}{k}\right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\lim\limits_{x\to\ 0}(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)})$
Evaluate $$\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)}\right)$$
It is easy with L'Hospital's rule, but takes too much time to calculate derivatives. What are some shorter methods?
| The series expansions are
$\sin x = x - x^3/6...,\,\,\tan x = x + x^3/3...,\,\, \arctan x=x - x^3/3...$
Equate numerators
$$\frac{(x+x^3/3)-(x-x^3/6)}{(x - x^3/6)(x + x^3/3)(x^3/3)}$$
$$\frac{x^3/2}{(x - x^3/6)(x + x^3/3)(x^3/3)}$$
We only care about the $x^3$ in the numerator
$$\frac{x^3/2}{x^3}$$
$$1/2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Remembering that $\sin^2(\theta) = 1/2 - 1/2\cos(2\theta)$? How do you remember this for integrals?
It doesn't seem obvious and I can never remember it when I come across it in integrals.
| $\cos^2 \theta + \sin^2 \theta = 1$
$\cos^2 \theta - \sin^2 \theta = \cos 2\theta$
Add and you get
$2\cos^2 \theta = 1 + \cos 2\theta$
$\cos^2 \theta = \frac 1 2 + \frac 1 2\cos 2\theta$
Subtract and you get
$2\sin^2 \theta = 1 - \cos 2\theta$
$\sin^2 \theta = \frac 1 2 - \frac 1 2\cos 2\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
Find a unit tangent vector to a curve that is an intersection of two surfaces. The intersection of the two surfaces given by the Cartesian equations $2x^2+3y^2-z^2=25$ and $x^2+y^2=z^2$ contains a curve $C$ passing through the point $P=(\sqrt{7},3,4)$. These equations may be solved for $x$ and $y$ in terms of $z$ to gi... | Hint a:
The normal to $2x^2+3y^2-z^2=25$ at $(\sqrt7,3,4)$ is parallel to the gradient: $(4x,6y,-2z)=2(2\sqrt7,9,-4)$
The normal to $x^2+y^2-z^2=0$ at $(\sqrt7,3,4)$ is parallel to the gradient: $(2x,2y,-2z)=2(\sqrt7,3,-4)$
The common tangent to both surfaces would be perpendicular to both of these normals, which woul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Coefficient Problem (polynomial expansion)
Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$
Just to begin,
$(1-x)(1+2x) = -2x^2 + x + 1$
$(1-x)(1+2x)(1-3x) = 6x^3 - 5x^2 - 2x + 1$
But expanding on like this take too long.
In the end th... | Somewhat more generally, let $$\prod_{j=1}^n (1 + (-1)^j j x) = 1 + c(n) x + d(n) x^2 + \ldots$$
so you want $d(15)$.
We have $c(0) = d(0) = 0$ with
$$ (1 + c(n-1) x + d(n-1) x^2)(1 + (-1)^n n x) = 1 + c(n) x + d(n) x^2 + \ldots $$
so that
$$ \eqalign{c(n) &= c(n-1) + (-1)^n n \cr
d(n) &= d(n-1) + (-1)^n n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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A limit question with a parameter $\displaystyle\lim_{a\to0} \int^{1+a}_0 \frac{1}{1+x^2+a^2}\,dx$
How to solve it?
Can I solve it in this way?
\begin{align}
&\lim_{a\to0} \int^{1+a}_0 \frac{1}{1+x^2+a^2}\,dx\\
&=\int^{1}_0 \lim_{a\to0} \frac{1}{1+x^2+a^2}\,dx\\
&=\int^{1}_0 \frac{1}{1+x^2}\,dx\\
&=π/4
\end{align}
The ... | $$1+a^2+x^2=(1+a^2)\left(1+\left(\frac{x}{\sqrt{1+a^2}}\right)^2\right)\implies$$
$$\int_0^{1+a}\frac{dx}{1+a^2+x^2}=\frac1{\sqrt{1+a^2}}\int_0^{1+a}\frac{\left(\frac1{\sqrt{1+a^2}}\right)dx}{1+\left(\frac{x}{\sqrt{1+a^2}}\right)^2}=$$
$$=\left.\frac1{\sqrt{1+a^2}}\arctan\frac x{\sqrt{1+a^2}}\right|_0^{1+a}=\frac1{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding the Range of a Trigonometric function The range of $$f(x)=3\cos^2x-8\sqrt3 \cos x\cdot\sin x+5\sin^2x-7$$is given by:(1)$[-7,7]$(2)$[-10,4]$(3)$[-4,4]$(4)$[-10,7]$
ANS: (2)
My Solution
The equation can be written as: $$3\cos^2x-8\sqrt3 \cos x\cdot \sin x+16\sin^2x-11\sin^2x-7 \\\implies (\sqrt3\cos x-4\sin x)^2... | The problem is when you say $y=0$ you fix the value of $sinx$ as $\sqrt3cosx-4sinx=0$ so $tanx=\frac{\sqrt3}{4}$ and $sin^2x= \frac{3}{19}$ not 0 or 1 . You should write it as the form of $asin2x+bcos2x+c$ as stated in the comment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Calculate $\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$ without using L'Hospital's rule Question:
Calculate
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$$
without using L'Hospital's rule.
Attempted solution:
First we multiply with the conjugate expression:
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} -... | $$
\sqrt{x^2+3x} - \sqrt{x^2+1} = x\sqrt{1+\frac{3}{x}} - x\sqrt{1+\frac{1}{x^2}} = x\left(\sqrt{1+\frac{3}{x}} -\sqrt{1+\frac{1}{x^2}} \right)
$$
expand the radicals we find
$$
\sqrt{1+\frac{3}{x}} = 1 + \frac{3}{2x} + O(x^{-2})\\
\sqrt{1+\frac{1}{x^2}} = 1 + O(x^{-2})
$$
put it all together
$$
\lim_{x\to\infty}\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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How does one calculate: $\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$ How does one calculate:
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$$
Is the best way to just take the first term times the following two, and the second two times the next two to see the pattern?
First ter... | Your series in parentheses is
$$\frac{1-\cos z} {z}. $$
Expand the binomial. The series for $\cos^2z$ is a bit complicated, but can be written using binomial coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calculating determinant of a Vandermonde type matrix of order n Det$
\begin{bmatrix}
1 & 2 & 3 &\ldots &n\\
1& 2^3& 3^3& \ldots & n^3\\
1 &2^5& 3^5& \ldots & n^5\\
\vdots & \vdots& & \vdots \\
1&2^{2n-1}& 3^{2n-1}& \ldots &n^{2n-1}
\end{bmatrix}
$
If the powers were consecutively increasing down the rows than we... | If you multiply the second column by $2$, the third by $3$ and so on, you get
$$
\det
\begin{bmatrix}
1 & 2 & 3 &\ldots &n\\
1& 2^3& 3^3& \ldots & n^3\\
1 &2^5& 3^5& \ldots & n^5\\
\vdots & \vdots& & \vdots \\
1&2^{2n-1}& 3^{2n-1}& \ldots &n^{2n-1}
\end{bmatrix}
=
\frac{1}{n!}\det
\begin{bmatrix}
1 & 2^2 & 3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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PDF of $Z=\frac{X^2+Y^2}{2}$ where $X\sim N(0,1)$ and $Y\sim N(0,1)$ Say $X \sim N(0,1)$ and $Y\sim N(0,1)$ are independent random variables. So: $f_X(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{-1}{2}x^2}$ and $f_Y(y) = \frac{1}{\sqrt{2\pi}}e^{\frac{-1}{2}y^2}$. Now I am interested in the probability density function (PDF) $f_... | $$\mathbb{P}[X^2+Y^2\leq R^2]=\frac{1}{2\pi}\iint_{x^2+y^2\leq r^2}e^{-\frac{x^2+y^2}{2}}\,dx\,dy=\frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^{R}\rho e^{-\rho^2/2}\,d\rho\,d\theta$$
so:
$$\mathbb{P}[X^2+Y^2\leq R^2]=1-e^{-R^2/2}$$
and $X^2+Y^2$ has an exponential distribution with parameter $\lambda=\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Sum of digits of $\dfrac{360}{2^n}$ always 9 - proper explanation? What is the exact explanation behind this "series"?
*
*$\dfrac{360} {2^1} = 180 => 1+8+0 = 9$
*$\dfrac{360}{2^2} = 90 => 9+0 = 9$
..
*$\dfrac{360}{2^{10}} => 0.3515625 => 0+3+5+1+5+6+2+5 = 27 => 2+7 = 9$
and so forth..
how is this kind of "rec... | Note that multiplying by $10$ doesn't change the sum of the digits, even if the digits extend beyond the decimal point, so the sum digits in $\frac{360}{2^n}$ is the same as the sum of digits of $\frac{360}{2^n} \times 10^n$, which is equal to $360 \times 5^n$. But
$$360 \times 5^n = 9 \times 40 \times 5^n$$
so this nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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One more confusing factoring question. The question is:
$x^6 + 5x^3 + 8$
Please can someone help me in factorising this. I saw some solutions but they are not meant for a IX grade student.
Thanks for the help.
| We can rearrange the polynomial as follows and use a difference of cubes:
\begin{align*}
x^6+5x^3+8&=(x^2+2)^3-6x^4+5x^3-12x^2\\
&=(x^2+2)^3-x^3-x^2(6x^2-6x+12)\\
&=\left[(x^2+2)-x\right]\left[(x^2+2)^2+x(x^2+2)+x^2\right]-6x^2(x^2-x+2)\\
&=(x^2-x+2)\left[(x^2+2)^2+x(x^2+2)-5x^2\right]\\
&=(x^2-x+2)\left(x^4+x^3-x^2+2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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What is $\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$?
What is $\displaystyle\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ ?
Find an asymptotic expansion of $\displaystyle \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ as $x\to 0$
One the one hand, $\sqrt{x}\to 0$, but $\displayst... | If we look at the corresponding integral, we get
$$\int_2^\infty \frac{\sqrt{x}\ln t}{1 + t^2x}\,dt = \int_{2\sqrt{x}}^\infty \frac{\ln u - \ln \sqrt{x}}{1+u^2}\,du.$$
Both functions,
$$\frac{\ln u}{1+u^2} \quad\text{and}\quad \frac{1}{1+u^2}$$
are Lebesgue integrable over $(0,\infty)$, so for the integral, we have th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finalising proof from Humphreys´ "Introduction to Lie Algebras and Representation Theyory" $L=\mathfrak{sl}(2, \mathbb{F})$ with standard Chevalley basis $(x, \ y, \ h)$ and $a, \ c\in \mathbb{Z}^{+}$. Humphrey gives a Lemma in chapter 26.2 saying:
$\frac{x^{c}}{c!}\frac{y^{a}}{a!}=\sum\nolimits_{k=0}^{min\{a,c\}} \fra... | You do need to keep careful track of the terms for the induction to go smoothly. In addition to the listed steps my calculation needed the formula (proof is straightforward)
$$
k\binom n k=(n-k+1)\binom n {k-1}.\qquad(*)
$$
I abbreviate
$$
y^{[k]}=\frac{y^k}{k!}
$$
and similarly with $x^{[k]}$. Things are a bit simple... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the value of $x$ such that $(3-\log_3x)\log _{3x}3=1$. Find the value of $x$ such that $(3-\log_3x)\log _{3x}3=1$.
Is there another way to solve other than this attempt?
My attempt,
$(3-\log_3x)\log _{3x}3=1$
$\frac{\log(3)\left(3-\frac{\log (x)}{\log (3)}\right)}{\log (3x)}=1$
$\log (3)\left(3-\frac{\log (x)}{\lo... | Using another approach, recall that $\log_b x=\frac{\log_c x}{\log_cb}$.
Now, for $b=3x$ and $c=3$ we have
$$\log_{3x}3=\frac{\log_33}{\log_3 3x}=\frac{\log_33}{\log_3 3+\log_3x}=\frac{1}{1+\log_3x}$$
Let $y=\log_3 x$. Then, the equation
$$\left(3-\log_3x\right)\log_{3x}3=1\implies\frac{3-y}{1+y}=1\implies y=1\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solve recurcion using generating function I have got a problem with solving this equation using generating functions.
$$ P_{n}=2nP_{n-1}-10n+5 $$
$$ P_{0}=5 $$
I started like that:
$$
f(x)=\sum_{n=0}^{\infty}P_{n}x^{n}=5+\sum_{n=1}^{\infty}P_{n}x^{n}=5+\sum_{n=1}^{\infty}(2nP_{n-1}-10n+5)x^{n}=5+\sum_{n=1}^{\infty}... | Method 1: Given $P_{n}=2nP_{n-1}-10n+5$ and $P_{0}=5$ then
\begin{align}
\sum_{n=0}^{\infty} P_{n+1} \, t^{n} &= 2 \, \sum_{n=0}^{\infty} (n+1) \, P_{n} \, t^{n} - 10 \, \sum_{n=0}^{\infty} n \, t^{n} - 5 \, \sum_{n=0}^{\infty} t^{n} \\
\sum_{n=1}^{\infty} P_{n} \, t^{n-1} &= 2 \, \frac{d}{dt} \, \sum_{n=0}^{\infty} P_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Finding the limit of $(1-\cos x)/x^2$ $$\lim _{x \to 0}{1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2$$
now $$\lim_{x \to 0}{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2... | I'm surprised nobody used L'Hospital's Rule:
$$\lim _{x \to 0}{1-\cos x\over x^2}\equiv \lim _{x \to 0}{\sin x\over 2x}\equiv\lim _{x \to 0}{\cos x\over 2}=\frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
$a^2+b$ and $a+b^2$ prime implies $\gcd(ab+1,a+b)=1$ Let $a,b>1$ be integers such that $a^2+b$ and $a+b^2$ are prime. Prove that $\gcd(ab+1,a+b)=1$.
Clearly $a$ and $b$ are of different parities; suppose $a$ is odd and $b$ even. If a prime $p\neq 2$ divides $ab+1$ and $a+b$, then it also divides $(ab+1)+(a+b)=(a+1)(b+1... | Continuing where you left off, suppose without loss of generality that $p$ divides $a+1$ and $b-1$. Then
$$
a+b^2 \equiv -1 + 1^2 \equiv 0 \pmod{p}
$$
Since $a+b^2$ is prime, it follows that $p=a+b^2$. But by assumption $p$ divides $a+b$, and $b > 1$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$xy + yz + zx + 2xyz = 1$ implies $4x+y+z\geq 2$ Let $x,y,z>0$ satisfy $$xy + yz + zx + 2xyz = 1.$$ Prove that $4x+y+z\geq 2$.
The condition invites the factoring $(1+x)(1+y)(1+z)+xyz-2=x+y+z$, but having the factor $4$ in the desired inequality makes things more difficult.
| since
$$xy+yz+xz+2xyz=1$$
Note this following indentity
$$\sum_{cyc}\dfrac{ab}{(b+c)(c+a)}+\dfrac{2abc}{(a+b)(b+c)(c+a)}=1$$
so we Let $$x=\dfrac{a}{b+c},y=\dfrac{b}{c+a},z=\dfrac{c}{a+b},a,b,c>0$$
$$\Longleftrightarrow \dfrac{4a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge 2\tag{1}$$
Use Cauchy-Schwarz inequality we have
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve the system of equations by variable estimation Solve the system of equations: $\left\{\begin{array}{l}(x-1)\sqrt{x-y^2}=y(x-2y+1)\\y\sqrt{x-1}+3\sqrt{x-y^2}=2x+y-1\end{array}\right.$
I guess there is only one solution $(x;y)=(2;1)$.
This is my try
Condition: $x\ge 1;x\ge y^2$.
We have: $x-2y+1\ge x-2|y|+1\ge x-... | since you have
$$x^2-x+2y^2-2xy=0\Longrightarrow x-y^2=x^2-2xy+y^2=(x-y)^2$$
Note $x>y$,so have
$$\sqrt{x-y^2}=x-y$$
take the second equation we have
$$y\sqrt{x-1}+3(x-y)=2x+y-1$$
so we have
$$y=\dfrac{x+1}{4-\sqrt{x-1}}\in[0,1]\Longrightarrow 1\le x\le 2$$
take
$$x^2-x+2y^2-2xy=0\Longrightarrow x^2-x+2\left(\dfrac{x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to proceed with evaluating $\int\frac{dx}{\sqrt{9+4x^2}}$ and $\int\tan^2(3x)dx$
*
*$\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}$
*$\displaystyle\int\tan^2(3x)dx$
For the first one i'm not sure if I did it correctly, here is what I did:
Let $2x=3\tan(t)$, so $x=\frac{3}{2}\tan(t)$ and $dx=\frac{3}{2}\sec^2(t)dt... | The first one is correct. For the second one, you can use the hopefully well-known derivative:
$$
D\tan x=1+\tan^2x.
$$
Hence,
$$
\int \tan^2(3x)\,dx=\int 1+\tan^2(3x)-1\,dx=\frac{1}{3}\tan(3x)-x+C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
calculate the $1/6+1/12+1/24+1/48 \ldots $. Wolfram is wrong? I am trying to calculate the following sum
$$
S = \frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48} + \cdots
$$
so
$$
S+\frac{1}{3} = \frac{1}{3} + \frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48} + \cdots = \frac{1}{3} \cdot \sum_{k=0}^{\infty} \frac{1}{... | Here's a link. Wolfram Alpha looks right to me. Your link is rather unclear, likely truncated: what exactly did you look up on Alpha?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Substituting a value of sine function in a trigonometric equation I am trying to really understand trigonometric equations and I've stumbled upon a rather confusing example. Solve the following equation:
$\sin x= 2|\sin x|+ {\sqrt{3}}\cos x$
First step is to define the absolute $\sin x$:
$$|\sin x| =
\begin{cases}
\... | Hint:
note that if $y=\dfrac{2 \pi}{3}$ than $\cos y = -\dfrac{1}{2}$, so you cannot use $\dfrac{1}{2}\sin x + \dfrac{\sqrt{3}}{2}\cos x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit? The limit is
$$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$
which I'm aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using t... | If we take one more term in the Taylor expansion:
\begin{align}
\sin x&\approx x-\frac{x^3}6+\cdots\\
\sin^2 x&\approx x^2-2x\frac{x^3}6+\cdots\\
\frac 1{\sin^2 x}&\approx\frac 1{x^2-x^4/3}\\
&=\frac 1{x^2}\cdot\frac 1{1-x^2/3}\\
\lim_{x\to 0}\left[\frac 1{\sin^2 x}-\frac 1{x^2}\right]&=\lim_{x\to 0}\left[\frac 1{x^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 3
} |
Help to solve this geometry problem
How to find the length of $BG$ and $EC$ using $\alpha$ and $\beta$ ?
| Writing
$$a := |\overline{BC}| \qquad b := |\overline{CA}| \qquad c := |\overline{AB}|$$
$$y := |\overline{CE}| = |\overline{ED}| \qquad z := |\overline{BG}| = |\overline{GF}|$$
we have, in right triangles $\triangle ADG$ and $\triangle AFE$,
$$\frac{b-2y}{c-z} = \cos A = \frac{c-2z}{b-y}$$
Solving for $y$ and $z$ gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\int \limits_0^{\infty} x^2 \exp(-2x^2) dx$ How to evaluate this integral?
$$\int \limits_0^{\infty} x^2 \exp(-2x^2) dx$$
I found similar problem, but don't know how to apply them here.
What do I have to substitute?
| There are often several ways to evaluate some integrals. Here is one of the less standard ways.
Consider the integral
\begin{align}
I(a) = \int_{0}^{\infty} x^2 \, e^{- a x^2} \, dx
\end{align}
for which
\begin{align}
I(a) &= - \partial_{a} \, \int_{0}^{\infty} e^{-a x^2} \, dx
= - \frac{1}{2} \, \partial_{a} \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Proof:Taylor expansion of inverse trigonometric functions I find it quite difficult to remember the Taylor expansion of inverse trigonometric functions.Actually in school we have been just taught the series (for finding limits in calculus without teaching us the proof).
Can someone provide a short proof of these expans... | You can remember the generalized binomial formula
$$(1+x)^p=1+px+\frac{p(p-1)x^2}2+\frac{p(p-1)(p-2)x^3}{3!}+\frac{p(p-1)(p-2)(p-3)x^4}{4!}\cdots,$$
and apply it for $p=-1$ (fairly easy)
$$\frac1{1+x}=1-x+x^2-x^3+x^4\cdots,$$
or $p=-1/2$,
$$\frac1{\sqrt{1+x}}=1-
\frac{x}2+
\frac{3\,x^2}{2^2\cdot 2}-
\frac{3\cdot5\,x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Minimum value of cosA+cosB+cosC in a triangle ABC I have used jensen's inequality but couldn't move on.
| $\cos A+\cos B+\cos C$
$=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2+1-2\sin^2\dfrac C2$
$=2\sin\dfrac C2\cos\dfrac{A-B}2+1-2\sin\dfrac C2\cos\dfrac{A+B}2$ as $A+B=\pi-C,\dfrac{A+B}2=\dfrac\pi2-\dfrac C2$
$=1+2\sin\dfrac C2\left[\cos\dfrac{A-B}2-\cos\dfrac{A+B}2\right]$
$=1+4\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$
Now $\sin\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Alternative Quadratic Formula Well the formula for solving a Quadratic equation is :
$$\text{If }\space ax^2+bx+c=0$$
then
$$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$
But looking at this : [Wolfram Mathworld] (And also in other places)
They give An Alternate Formula:
$$x=\dfrac{2c}{-b \pm \sqrt{b^2 -4ac}}$$
How does one... | \begin{align}
\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}
&= \dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}
\cdot \frac{-b \mp \sqrt{b^2 -4ac}}{-b \mp \sqrt{b^2 -4ac}} \\
&= \dfrac{b^2-(b^2-4ac)}{2a(-b \mp \sqrt{b^2 -4ac})} \\
&= \dfrac{4ac}{2a(-b \mp \sqrt{b^2 -4ac})} \\
&= \dfrac{2c}{-b \pm \sqrt{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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The value of $x$ for which function attains max value
At what value of $x,\ x\in \mathbb{Z}$ will the function $\dfrac{x^2+3x+1}{x^2-3x+1}$ attain its maximum value .
$\color{green}{a.)\ 3 }\\
b.)\ 4 \\
c.) -3 \\
d.)\ \text{none of these} \\ $
$\dfrac{x^2+3x+1}{x^2-3x+1}\\
=1+\dfrac{6x}{x^2-3x+1}\\
=1+\dfrac{6}{x-3+\... | Your last two formulas for $f(x)$ in your first edit are incorrect: they should be
$$f(x)=1+\frac{6x}{x^2-3x+1}=1+\frac{6}{x-3+\frac{1}{x}}$$
(I see you corrected that formula in your question by a later edit.)
You maximize that by making the denominator positive but as small as possible. The $-3$ in that denominator m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Given primitive solution to $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$, show $a+b$ is a perfect square If $a,b,c$ are positive integers and
$\gcd(a,b,c)$ is $1$. Given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ then prove that $a+b$ is a perfect square.
I was trying to get something useful from the information given in... | Ah well, since Jack deleted his answer, I'll fix it for him.
The equation is equivalent to $ac+bc=ab$ or $(a-c)(b-c)=c^2$.
Claim: $\gcd(a-c,b-c)=1$.
Proof: $\gcd(a,b,c)=1$ implies $\gcd(a-c,b-c,c)=1$ which implies $\gcd(a-c,b-c,c^2)=1$. But $a-c\mid c^2$, so $\gcd(a-c,b-c,c^2)=\gcd(a-c,b-c)$.
When the product of two ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Find the last two digits of $33^{100}$
Find the last two digits of $33^{100}$
By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$
So $33^{40}\equiv 1 \pmod{100}$
Then how to proceed?
With the suggestion of @Lucian:
$33^2\equiv-11 \pmod{100}$ t... | $$\begin{align}
33^{100}&=9^{50}\cdot11^{100}\\
&=(1-10)^{50}(1+10)^{100}\\
&=(1-50\cdot10+\cdots)(1+100\cdot10+\cdots)\\
&\equiv1\mod100
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ where $\omega^3 = 1$ If $\omega^3 = 1$ and $\omega \neq 1$, then compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$
I'm pretty lost, I don't really know where to start.
Thanks
| Note that $x^3-1=(x-1)(x^2+x+1)$ and $\omega\neq 1$ so that $\omega^2+\omega+1=0$
This means that $1+\omega-\omega^2=-2\omega^2$ and $1-\omega+\omega^2=-2\omega$
The product then reduces to $4\omega^3=4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Complex numbers - roots of unity
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}.$$
I have tried adding the first two and the second two separately, then ad... | A less efficient answer with a more general approach.
You may notice that $\{\omega,\omega^2,\omega^3,\omega^4\}$ are the roots of $\frac{x^5-1}{x-1}$. If we set $Z=\{\omega,\omega^2,\omega^3,\omega^4\}$
we have
$$ \sum_{z\in Z}\frac{z}{1-z^2}=\sum_{z\in Z}\frac{z^3}{1-z}=\sum_{z\in Z}\frac{1}{1-z}-\sum_{z\in Z}(1+z+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Matrix multiplication and determinant question Show that if $\det(\begin{bmatrix}b & c\\a & b\end{bmatrix})=0$ with $A=\begin{bmatrix}a & a\\b & b\end{bmatrix}$ and $B=\begin{bmatrix}b & b\\c & c\end{bmatrix}$ then $AB=BA.$
How do I go by solving this? I tried finding the determent, $AB$, and $BA$ but it didn't work.
| $$
\begin{bmatrix} a & a \\ b & b \end{bmatrix} \begin{bmatrix} b & b \\ c & c \end{bmatrix} = \begin{bmatrix} ab+ac & ab+ac \\ c^2+bc & b^2 + bc \end{bmatrix}
$$
$$
\begin{bmatrix} b & b \\ c & c \end{bmatrix} \begin{bmatrix} a & a \\ b & b \end{bmatrix} = \begin{bmatrix} ab+b^2 & ab+b^2 \\ ac+bc & ac + bc \end{bmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the coefficient of $x^9$ in $(1+x)(1+x^2)(1+x^3)\cdots(1+x^{100})$ This question had come in jee advanced 2015. Give a hint to solve it.
| You have $100$ factors, each a sum of a $1$ and a positive power of $x$. When the product is expanded as a sum, every term results from picking a $1$ from some factors and a positive power of $x$ from others. If the positive powers of $x$ that you pick in one case are $x^3$, $x^{20}$, and $x^{35}$, then the term is $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find the residue of $e^{\frac{1}{z^2-1}}\sin(\pi z)$ at $z=1$ I'm dealing with the following problem (from an old qualifying exam):
Let $\gamma$ be a closed curve in the right half-plane that has index $N$ with respect to the point 1. Find
$$
\int_{\gamma}e^{\frac{1}{z^2-1}}\sin(\pi z)\,dz.
$$
I take this problem a... | Expanding the functions $\exp$ and $\sin$ in Laurent series we get
[
\begin{align}
\exp \left( {\frac{1}{{z^2 - 1}}} \right) = \sum {\frac{1}{{\left( {z^2 - 1} \right)^n n!}}} = \sum {\frac{1}{{\left( {z + 1} \right)^n \left( {z - 1} \right)^n n!}}} = \sum {\frac{{\frac{1}{{\left( {z + 1} \right)^n }}}}{{\left( {z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the inverse Laplace transform of: $\frac{1}{(s^2+a^2)(s^2+b^2)}$ I'm having trouble doing this homework problem because I'm not sure how to deal with the $a$ and $b$. I did it the usual way we were taught - use partial fraction decomposition and then try to solve for the coefficients. When I solved for them, this ... | Just note $$\frac{1}{s^2+a^2}-\frac{1}{s^2+b^2}=\frac{b^2-a^2}{(s^2+a^2)(s^2+b^2)}$$
Then, if $a^2\neq b^2$ and $ab\neq 0$, we have
\begin{align*}
\frac{1}{(s^2+a^2)(s^2+b^2)}&=\frac{\frac{1}{b^2-a^2}}{s^2+a^2}-\frac{\frac{1}{b^2-a^2}}{s^2+b^2}\\
\mathcal{L}^{-1}\left\{\frac{1}{(s^2+a^2)(s^2+b^2)}\right\}&=\frac{1}{a(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the value of $x$ such that $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$ Find the value of $x$, $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$$
Help guys please, I have tried and I got, $x=-2, x=1$, and I think it's wrong
| $$\begin{align}\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}&=x&\Longleftrightarrow \\
\left(\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}\right)^2&=x^2&\Longleftrightarrow\\
4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}&=x^2&\Longleftrightarrow\\
\left(\sqrt{4-\sqrt{4+\sqrt{4-x}}}\right)^2&=\left(x^2-4\right)^2&\Longleftrightarrow\\
4-\sqrt{4+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Finding the roots of a different Quadratic equation from the roots of a Given Quadratic equation The Question:
If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$...
Then find the roots of the equation $ax^2-bx(x-1)+c(x-1)^2=0$
My Attempt:
The new equation can be made into a quadratic as:
$$(a-b+c)x^... | hint: $\dfrac{2c-b}{a-b+c} = \dfrac{2\dfrac{c}{a}-\dfrac{b}{a}}{1-\dfrac{b}{a}+\dfrac{c}{a}}=\dfrac{2\alpha\cdot \beta+(\alpha+\beta)}{1+\alpha+\beta+\alpha\cdot \beta}$, can you continue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving basic floor function inequality: $-1 \lt \lfloor 2x \rfloor - 2 \lfloor x \rfloor \lt 2$ As a direct consequence of the definition of $\lfloor x \rfloor $ I know that $$2x-1 \lt \lfloor 2x \rfloor \le 2x$$ and $$2x-2 \lt 2\lfloor x\rfloor \le 2x$$
How can I use these to show that $-1 \lt \lfloor 2x \rfloor - ... | Let $n = \lfloor x\rfloor$ and $f=x-n = x-\lfloor x\rfloor$, i.e. $x=n+f$. Then
$$\begin{align*}
\lfloor 2x\rfloor-2\lfloor x\rfloor&= \lfloor 2n+2f\rfloor - 2\lfloor n+f\rfloor\\
&=2n+\lfloor 2f\rfloor - 2n\\
&= \lfloor 2f\rfloor
\end{align*}$$
Since $0\le f<1$, $0\le 2f<2$ and then $\lfloor 2f\rfloor$ can only be $0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Side limits of the derivative of this function $f:\mathbb{R}\to \mathbb{R}$ with $f\left(x\right)=\left(x^3+3x^2-4\right)^{\frac{1}{3}}$
Calculate side limits of this function's derivative, $f'_s\:and\:f'_d$, in $x_o=-2$
The answer key says I should get $\infty $ and $-\infty$ but I'm not getting that. The derivative I... | You have the wrong expression for the derivative.
$$\begin{align}
\frac d{dx}\left[(x^3+3x^2-4)^{1/3}\right]
&= \frac 13(x^3+3x^2-4)^{-2/3}(3x^2+6x) \\[2ex]
&= \frac {x^2+2x}{[(x+2)^2(x-1)]^{2/3}} \\[2ex]
&= \frac {x(x+2)}{(x+2)^{4/3}(x-1)^{2/3}} \\[2ex]
&= \frac {x}{(x+2)^{1/3}(x-1)^{2/3}} \\[2ex]
\end{align}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find eigenvalues and eigenvectors of this matrix Problem: Let \begin{align*} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}. \end{align*} Compute all the eigenvalues and eigenvectors of $A$.
Attempt at solution: I found the eigenvalues by computing the characteristic ... | You are exactly right! $x_1,x_2,x_3$ and $x_4$ can be anything as long as they sum to zero. This is already a complete solution in some sense (you have "found" all of the eigenvectors). If you would like, you could find 3 linearly independent such vectors, and then you would be sure that these three span the whole s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $LK_1^2 + LK_2^2 + \dots + LK_{11}^2$. $K_1 K_2 \dotsb K_{11}$ is a regular $11$-gon inscribed in a circle, which has a radius of $2$. Let $L$ be a point, where the distance from $L$ to the circle's center is $3$. Find
$LK_1^2 + LK_2^2 + \dots + LK_{11}^2$.
Any suggestions as to how to solve this problem? I'm unsu... | Let $\omega = e^{2 \pi i/11}$, a primitive $11^{\text{th}}$ root of unity. We can assume that the circle is centered at the origin. We can also assume that $A_k$ is associated with the complex number $2 \omega^k$
Let $p$ be complex number associated with the point $P$. Then
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Way to make this homogoneous ODE seperable? Is my algebra correct, turning this:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4y-3x}{2x-y}$$
Into this:
I split into the difference of the two fractions,
then factored x out of the left fraction, and factored y
out of the right fraction, getting:
$$\frac{4\frac{y}{x}}{x(1-\... | I solved it maybe it will help you:$$\frac { dy }{ dx } =\frac { 4y-3x }{ 2x-y } =\frac { 4\frac { y }{ x } -3 }{ 2-\frac { y }{ x } } \\ \frac { y }{ x } =t$$
$$ y=xt\\ y^{ \prime }=t+x{ t }^{ \prime }\\ t+x{ t }^{ \prime }=\frac { 4t-3 }{ 2-t } $$ $$ x{ t }^{ \prime }=\frac { { t }^{ 2 }+2t-3 }{ 2-t } =\frac { \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Using equation to find value of $1/x - 1/y$ $$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
What is the value of $\frac{1}{x}-\frac{1}{y}$?
I have already used that when $48$ divided by $10$ then it becomes $4.8$ and when $8$ divided by $10$ then it becomes $0.8$ by getting $10^x$ and $10^y$ to make i... | From $4.8^x = 1000$ we get $4.8 = 1000^{1/x}$ and similarly for the $y$ term we get $0.8 = 1000^{1/y}$ by taking the $x$ and $y$-root of both sides respectively. Hence we get $$\frac{4.8}{0.8} = \frac{1000^{1/x}}{1000^{1/y}}$$
This yields $$6 = 1000^{1/x - 1/y}$$ so $$\bbox[10px, border: solid red 2px]{\frac{1}{x} - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Limit and factorization I have the following very interesting homework exercise:
Let $$f(x)=\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}$$ Find the
following limit, if it exists: $$\lim_{x\to 2}f(x)$$
I understand that I need to "delete" the $x-2$ factor from both nominator and denominator and then evaluate the li... | $$f(x)=\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}\Longrightarrow$$
$$\lim_{x\to2}f(x)=\lim_{x\to2}\left(\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}\right)=$$
$$\lim_{x\to2}\left(\frac{\frac{d}{dx}\left(x^2-4\right)}{\frac{d}{dx}\left(2-x\cdot \sqrt {x+2}+\sqrt{x+2}\right)}\right)=$$
$$\lim_{x\to2}\left(\frac{2x}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Verify that $\binom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$ for $n \geq 4$
Verify that for $n \geq 4$
$$\dbinom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$$
Now present a combinatoric argument for the above.
First, by verify does it mean check for so... | Using
\begin{align}
\binom{n}{2} = \frac{n(n-1)}{2}
\end{align}
then
\begin{align}
\frac{1}{3} \, \binom{\binom{n}{2}}{2} &= \frac{1}{2} \binom{n}{2} \, \left(\frac{n}{2} - 1\right) \\
&= \frac{n(n-1)}{4!} \left( n(n-1)-2 \right) = \frac{(n+1)(n)(n-1)(n-2)}{4!} \\
&= \binom{n+1}{4}.
\end{align}
As to the modified c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Does $(a^p + b^p)^{p-1} \equiv 1 \pmod {p^2}$ have any solutions where $a$ and $b$ are co-primes less than $p$? How will you prove that $(a^p + b^p)^{p-1} \equiv 1 \pmod {p^2}$ has no solution where $p$ is a prime number and $a$, $b$ are two co-primes less than $p$? If this equation has a solution, then what it is it?
... | For $p=2$ it says $a^2 + b^2 \equiv 1 \mod 4$. This can only be achieved if one of $a$ and $b$ is divisible by 2, so this one doesn't have any solutions.
For $p=3$, one has $(a^3+b^3)^2 \equiv 1 \mod 9$, so $a^3+b^3 \equiv 1 \mod 9$ or $a^3+b^3 \equiv 8 \mod 9$. The former has no solutions, same argument as for $p=2$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $2xy$ .
If $13x+17y=643$ ,$\{x,y\}\in \mathbb{N}$, then what is the value of two times the product of
$x$ and $y$ ?
Options
$a.)\ 744\quad \quad \quad \quad \quad
b.)\ 844\\
\color{green}{c.)\ 924}\quad \quad \quad \quad \quad
d.)\ 884\\$
I tried,
$13x+17y \pmod{13}\equiv 0\\
\implies 2y \pm... | When applying mod $13$, the equation $13x+17y=643$ becomes
$$4y\equiv 6\pmod{13}$$
or
$$40y\equiv 60\pmod {13}$$
that is, $y\equiv 8\pmod {13}$.
Now, to find $x$, apply mod $17$:
$$13x\equiv 14\pmod{17}$$
or
$$4\cdot 13x\equiv 56\pmod {17}$$
thus, $x\equiv 5\pmod{17}$.
Now we are to find the concrete values of $x$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Why does this sum converge? I know that the following sum converges to 2 via WolframAlpha, but I am not sure why.
$$\sum_{k=1}^\infty k \left[\frac{2}{k} - \frac{4}{k+1} + \frac{2}{k+2}\right] = 2$$
WolframAlpha gives the following partial sum formula:
$$\sum_{k=1}^n k \left[\frac{2}{k} - \frac{4}{k+1} + \frac{2}{k+2... | $$\begin{align}k\left(\frac 2k-\frac{4}{k+1}+\frac{2}{k+2}\right)&=2-\frac{4k}{k+1}+\frac{2k}{k+2}\\&=2-4\left(1-\frac{1}{k+1}\right)+2\left(1-\frac{2}{k+2}\right)\\&=4\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Derive Cartesian cubic Möbius strip from parametric The following link:
http://mathworld.wolfram.com/MoebiusStrip.html
shows the Möbius strip parametrized as
\begin{eqnarray}
x = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \cos t \\
y = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \sin t \\
z = s \sin \left ( \frac1... | Consider the Moebius strip with midline the circle of radius $R$ in the $(x,y)$-plane and having width $2a>0$:
$$M:\quad(\phi,s)\mapsto\left\{\eqalign{x&=\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\cos\phi \cr
y&=\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\sin\phi \cr
z&=s\sin{\textstyle{\phi\over2}}\ .\cr}\right.\tag{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the principal solutions of the trigonometric equation $\cos x-\sin x+\sin 2x+3\cos2x+1=0$ I am unable to simplify the expression. If I simplify the double angles, it leaves me with a nasty expression,
$\cos x-\sin x+2\sin x\cos x+6\cos^2 x-2=0$. What do I do next. Some hints, please. Also, is there some elegant s... | Observe:
$$
\cos x-\sin x+2\sin x\cos x+6\cos^2 x-2=(\sin x)(2\cos x-1)+(2\cos x-1)(3\cos x+2)=(\sin x+3\cos x-2)(2\cos x-1).
$$
Therefore, $\cos x=\frac{1}{2}$ is a solution.
The solutions to $\sin x+3\cos x+2=0$ are less pleasant, but one can solve $\sin x=-2-3\cos x$, square both sides and solve, using the quadratic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Volume of Tetrahedron $ABCD$ is a regular tetrahedron of volume $1$. Maria glues regular tetrahedra $A'BCD$, $AB'CD$, $ABC'D$, and $ABCD'$
to the faces of $ABCD$. What is the volume of the tetrahedron $A'B'C'D'$?
| Let the edge length of (original) regular tetrahedron $ABCD$ (centered at $O$) be $a$ then the volume of the regular tetrahedron $ABCD$ $$=\color{red}{\frac{a^3}{6\sqrt{2}}}$$ but the volume is $1$ hence, we have $$\frac{a^3}{6\sqrt{2}}=1$$ $$\implies \color{blue}{a=(6\sqrt{2})^{1/3}}$$ Distance of each face of regular... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solving $6 \cos x - 5 \sin x = 8$ My attempt:
Using the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi)$$
$$
\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8
$$
$$
\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}
$$
And... | We have $$6 \cos x - 5\cos x = \cos x$$
But, a quick look at the graph of the cosine function shows us that it is bounded between $-1$ and $1$, so $\cos x = 8$ has no solutions. $\square$
Assuming your original equation was $6 \cos x - 5\sin x = 8$ as your body would suggest, we can represent this in the form $$\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Wanted : for more formulas to find the area of a triangle? I know some formulas to find a triangle's area, like the ones below.
*
*Is there any reference containing most triangle area formulas?
*If you know more, please add them as an answer
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=... | If $W$ is the "hypotenuse-face" of a right-corner tetrahedron, and $X$, $Y$, $Z$ are the (right-triangular) "leg-faces", then
$$W^2 = X^2 + Y^2 + Z^2$$
where, yes, we are squaring areas. (This fact is actually equivalent to Heron's formula for non-obtuse triangles. You can extend it to include obtuse triangles by all... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 3
} |
$ x^2 + \frac {x^2}{(x-1)^2} = 2010 $ I found this question from last year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Given $$ x^2 + \frac {x^2}{(x-1)^2} = 2010,$$
find $\dfrac {x^2} {x-1}.$
(A) $1+\sqrt {2011}$ (B) $ 1-\sqrt {2011}$ (C) $1\pm \sqrt{2011} $... | A straightforward approach not requiring recognition of an identity runs:
Let $y=\frac {x^2}{x-1}$ so that $x^2=y(x-1)$, which means we can reduce $x^2$ wherever it occurs and see what happens. So $$x^2+\frac {x^2}{(x-1)^2}=y\left(x-1+\frac 1{x-1}\right)=y\cdot\frac {x^2-2x+2}{x-1}=y^2-2y$$And it is easy from there (so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $x$ if $\frac {1} {x} + \frac {1} {y+z} = \frac {1} {2}$ I found this question from past year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Find $x$ if \begin{align}\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac... | Let $x=1/p$, $y=1/q$, and $z=1/r$. The three equations become
$$\begin{align}
{1\over2}&=p+{qr\over q+r}={s\over q+r}\\
{1\over3}&=q+{rp\over r+p}={s\over r+p}\\
{1\over4}&=r+{pq\over p+q}={s\over p+q}\\
\end{align}$$
where
$$s=pq+qr+rp$$
Thus
$$\begin{align}
2s&=q+r\\
3s&=r+p\\
4s&=p+q\\
\end{align}$$
From this it f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Factorisation over $\Bbb C$ of $z^2 -10z+30$ I haven't done these questions in a long time, so I am just wondering if my approach and answer is correct.
When asked to $z^2-10z+30$ over $\Bbb C$,
My approach: I complete the square of the equation, and would get $(z-5+\sqrt 5)(z-5-\sqrt 5)$
My question/concern: I think ... | The "$i$" should be included, thusly:
$z^2 - 10z + 30 = (z - 5 + i\sqrt{5})(z - 5 - i\sqrt{5}) ; \tag{1}$
without it, we have
$(z - 5 + \sqrt{5})(z - 5 - \sqrt{5}) = ((z - 5) + \sqrt{5})((z - 5) -\sqrt{5})$
$= (z - 5)^2 - (\sqrt {5})^2 = z^2 - 10z + 25 - 5$
$= z^2 - 10z + 20; \tag{2}$
on the other hand, with it:
$(z -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Simultaneous equations, $\frac{1}{x}+\frac{1}{y}=1$,$x+y=a$,$\frac{y}{x}=m$ By eliminating $x$ and $y$ from the following equations, I need to find the relation between $m$ and $a$.
\begin{align*}
\frac{1}{x}+\frac{1}{y}=1 \\
x+y=a \\
\frac{y}{x}=m
\end{align*}
I tried different ways, but cannot arrive at the answer. ... | No, since for each value of $y$ there is at most one $x$ satisfying $x+y=a$. You can now reexamine your solution.
Here is a different approach:
From the last two equations, we can derive that $x=\frac{a}{1+m}$ and $y=\frac{ma}{1+m}$. Substitute into the first equation, and we have $(1+m)^2=ma$. It is clear $m\ne0$. H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h $, find $f(7)$. Problem :
Given a polynomial $f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h$
such that $f(1)= 1, f(2) =2 , f(3) = 3, f(4) =4, f(5)=5, f(6) =6$.
Find $f(7)$ in terms of $h$.
My approach:
We can put the values of $f(1) = 1$ in the given equation and $f(2) = 2$, etc. But ... | \begin{align}
f(x) &=ax^6 +bx^5+cx^4+dx^3+ex^2+fx+3
\end{align}
Given that :
$f(1)= 1$, $f(2) =2$ , $f(3) = 3$, $f(4) =4$, $f(5)=5$, $f(6) =6$,
find $f(7)$.
It is also useful to add that $f(0)=3$.
Application of successive differences
lowers the degree of the polynomial on every step
down to $0$, and this useful formul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 3
} |
Proving $\sum_{i=1}^n\frac{1}{i(i+1)(i+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ for $n\geq 1$ by mathematical induction Prove using mathematical induction that
$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \cdots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}.$$
I tried taking $n=k$,
so it makes
$$\frac{1}{1\... | Given that $$\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+\ldots +\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$$
1. Substituting $\color{blue}{n=1}$ in the given equality, we get $$\frac{1}{1\times 2\times 3}=\frac{(1)(1+3)}{4(1+1)(1+2)}$$ $$\implies \frac{1}{6}=\frac{4}{24}\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Infinite limit of trigonometric series The value of $\displaystyle\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))$ is
(A) $\sin^4x$
(B) $\sin^2x$
(C) $\cos^2x$
(D) does not exist
My attempt:
$$\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))=$$
$$=(\sin... | $$\begin{align} \sin^2 (x) - \sin^4(x) &= \sin^2 (x) \cos^2(x) = \frac14 \sin^2(2x), \\
\frac14 \sin^2 (2x) - \frac14 \sin^4(2x) &= \sin^2 (2x) \cos^2(2x) = \frac{1}{4^2} \sin^2(4x),\\
&\cdots \\
\frac{1}{4^{n}} \sin^2 (2^{n}x) - \frac{1}{4^{n}} \sin^4(2^{n}x) &= \sin^2 (2^{n}x) \cos^2 (2^{n}x)= \frac{1}{4^{n+1}} \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
| Rearranging, we have
$$\frac{1}{1+(a+b)}+\frac{1}{1-(a+b)}+\frac{1}{1-(a-b)}+\frac{1}{1+(a-b)},$$
giving
$$\frac{2}{1-(a+b)^2}+\frac{2}{1-(a-b)^2}.$$
Then,
$$\frac{4-2(a-b)^2-2(a+b)^2}{1-(a-b)^2-(a+b)^2+(a+b)^2(a-b)^2}.$$
Expanding,
$$\frac{4(1-a^2-b^2)}{1-2a^2-2b^2+(a^2-b^2)^2}.$$
In your case
$a=\sqrt{2}$ and $b=\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Summation of a logarithmic series for $\ln(2(r^2 - 1)/r^2)$ Given that
$$\sum_{r=2}^{n}\ln\frac{r^2-1}{r^2}=\ln\frac{n+1}{2n}$$
for $n >1$.
Express $$\sum_{r=32}^{62}{\ln\frac{2(r^2-1)}{r^2}}$$ as $$A\ln 2 + B\ln3 + C\ln7$$ where $A$, $B$, $C$ are positive integers that can be found.
The result is actually $\ln 2113929... | You don't need to deal with huge numbers.$$\begin{align}\sum_{r=32}^{62}\ln\left(\frac{2(r^2-1)}{r^2}\right)&=\sum_{r=32}^{62}\left(\ln 2+\ln\left(\frac{r^2-1}{r^2}\right)\right)\\&=\sum_{r=32}^{62}\ln 2+\sum_{r=32}^{62}\ln\left(\frac{r^2-1}{r^2}\right)\\&=(62-32+1)\ln 2+\sum_{r=2}^{62}\ln\left(\frac{r^2-1}{r^2}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finite difference : relationship involving gamma Given the following PDE,
$$
\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2V}{\partial S^2}=0
$$
and its finite difference approximation,
$$
\frac{V_n^{m+1}-V_n^m}{\Delta t} + \frac{1}{2}\sigma^2S_n^2\Gamma_n^m=0 \qquad (*)
$$
where
$$
\Gamma_n^m... | Basically we need to show that
$$
D (S_n^2 \Gamma_n) = S_n^2 D \Gamma_n + 4S_n \frac{\Gamma_{n+1} - \Gamma_{n-1}}{2\Delta S} + \Gamma_{n+1} + \Gamma_{n-1}
$$
where $D$ is second order finite difference derivative operator
$$
D u_n \equiv \frac{u_{n+1} -2 u_n+ u_{n-1}}{\Delta S^2}
$$
Note, that $D$ can be written as
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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.