Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find a and b such that $x^2 + 8x + 7 = (x + a)(x + b)$ for all real values of $x$. Find $a$ and $b$ such that $x^2 + 8x + 7 = (x + a)(x + b)$ for all real values of $x$.
That is what the question states, I think you should factor
Any Ideas on how to begin?
|
Use $$x=\frac{-b\pm \sqrt{b^2 - 4ac}}{2a},\quad \textrm{where }ax^2+bx+c=0$$
which gives $x=-1 \textrm{ or } -7$ , so factorize will be $(x+1)(x+7)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all integer solutions to the equation $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$ I was going through one of my Mathematics books and I came to this problem:
Find all integer solutions to the equation: $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$
I tried a few things to begin with but none went in the right directi... | *
*First, let us suppose that $x,y,z$ are positive integers satisfying the proposed equality. Then
$$ xyz(15-x-y-z)=\frac{1}{2}\left((xy-yz)^2+(yz-zx)^2+(zx-xy)^2\right)\ge0$$
So we have $x+y+z\le 15$.
*Now, note that squares modulo 5 are $\{0,-1,1\}$. Thus, if the sum of three squares modulo $5 $ is $0$ then either ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Combinations and Summations" $\sum_{x=0}^{n/k} \binom n{kx}$ How can be calculate this following summation efficiently (is there some shorter formula for it) ?
$$\sum_{x=0}^{\frac{n}{k}}{n\choose k\cdot x}$$
| $$\sum_{k=0}^{\frac{n}{x}}\binom{n}{x\cdot k} = \binom{n}{x\cdot 1}+\binom{n}{x\cdot 2}+\binom{n}{x\cdot 3}+...+\underbrace{\binom{n}{x\cdot \frac{n}{x}}}_{\binom{n}{n}}$$
$\forall y \in \mathbb{N} : y = \frac{n}{x}$
$$\sum_{k=0}^{n}\binom{n}{x\cdot k} = \binom{n}{x\cdot 1}+\binom{n}{x\cdot 2}+\binom{n}{x\cdot 3}+\unde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding a function given a tangent to the curve I have got the following maths problem:
In the curve $y=x^2+ax+b$ where $a$ and $b$ are constant.
The tangent to the curve where $x=1$ is $2x+y=6$. Find the values of $a$ and $b$.
I am just unsure how I would go about answering this.
| Since the involved function is just a quadratic function, you can try not to use any calculus.
The question can be translated as follows:
The equation $x^2+ax+b= 6-2x$ has a double root at $x=1$,
or even $f(x)=x^2+(a+2)x+(b-6)$ can be factorized as $f(x)=k(x-1)^2$, by factor theorem.
Edit Seems my answer above lead t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all values of $c$ such that solutions of $x^2 + cx + 6 = 0$ are rational. This is what I've done so far:
Using the formula, we get
$x = \frac{-c \pm \sqrt{c^2 - 24}}{2}$
For solutions to exist, $c \geq 5$.
For solutions to be rational, $c^2 - 24$ has to be a perfect square.
I'm kind of stuck now. I'd appreciate if... | Observe that if $x \in \mathbb{Q}$ which is a solution of the equation $x^2 + cx + 6 = 0\text{ } (1)$, then solve for $c$ in terms of $x$ we obtain: $c = -\dfrac{x^2+6}{x}$. Thus if $c = - \dfrac{r^2+6}{r}$ with $r \in \mathbb{Q}\setminus \{0\}$, we prove the solutions of $(1)$ are rational numbers. We have: $x^2+ cx +... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prime numbers of the form $(1\times11\times111\times1111\times...)-(1+11+111+1111+...)$ Let
$$R(1) = 1-1,$$
$$R(2) = (1\times11) - (1+11),$$
$$R(3) = (1\times11\times111) - (1+11+111),$$ and so on...
$$R(4)=1355297\quad\text{(a prime number!)}$$
$R(4)$ is the only prime I found of such form up to $R(200)$. Are there an... | I wanted to post this as a comment but it was too long.
I don't know if it will help but consider to rewrite your function in this way:
$$R(n) = (1 \times 11 \times 111 \times \ldots) - (1 + 11 + 111 + 1111 + \ldots)$$
$$R(n) = \left((10^0) \times (10^0 + 10^1) \times (10^0 + 10^1 + 10^2) \times \ldots\right) - \left((... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 3
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Does not algebraic multiplicity= geometric multiplicity $\Rightarrow$ the matrix is diagonalizable?
let $A=\left(\begin{array}{ccc} -5 & -1 & 6 \\ -2 & -5 & 8 \\ -1 & -1 & 1 \end{array}\right)$ and $B=\left(\begin{array}{ccc} -9 & 3 & -3 \\ -14 & 4 & -7 \\ -2 & 1 & -4 \end{array}\right)$ matrices above $\mathbb{C}$ ar... | A well known theorem state that: two matrices are similar if and only if they have the same Jordan normal form . In your case, the Jordan normal forms are different, so they cannot be similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Numbers $p-\sqrt{q}$ having regular egyptian fraction expansions? I remind that the greedy algorithm for egyptian fraction expansion for a positive number $x_0 <1$ goes like this:
$$x_0=\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\dots$$
$a_n$ are positive integers and are defined:
$$x_n-\frac{1}{a_n}>0$$
$$x_n-\frac{1}{... | Suppose $u>1$. Then the numbers $c_n := u^n - u^{-n}$
satisfy the linear recurrence
$$
c_{n+1} - (u+u^{-1}) c_n + c_{n-1} = 0.
$$
Moreover,
$$
\frac1{c_n} = \frac{u^n}{u^{2n}-1} = \frac1{u^n-1} - \frac1{u^{2n}-1}.
$$
Hence the sum of the reciprocals of the $2^m$-th terms can be evaluted
as a telescoping sum:
$$
\sum_{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Why are only $n-1$ convergents solution to Pell's Equations Given any equation of form $x^2-ny^2 = 1$, where $n$ is positive, non-square integer, let $[a_0; \overline{a_1, a_2, \dots a_k}]$ denote the continued fraction $\sqrt{n}$.
Let $\frac{p_i}{q_i} = [a_0; a_1, a_2, \dots a_{(k-1)i}]$. Why is it true that the solut... | If $x,y$ are positive integers satisfying $x^2 - ny^2 = 1$, then $\frac{x}{y}$ is a good rational approximation to $\sqrt{n}$. A really good rational approximation:
\begin{align}
&& x^2 - ny^2 &= 1 \\
&\iff& \frac{x^2}{y^2} - n &= \frac{1}{y^2} \\
&\iff& \frac{x}{y} - \sqrt{n} &= \frac{1}{y^2\bigl(\sqrt{n} + \frac{x}{y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $ \begin{cases} z^{8}=|z|^{7} \\ z=\bar z e^{i\frac{\pi }{2}} \end{cases}$ someone can help me with the following system:
$$ \begin{cases}
z^{8}=|z|^{7} \\
z=\bar z e^{i\frac{\pi }{2}}
\end{cases}$$
I found the solution $z=0$
In fact from the second equation:
$$z=i\bar z$$
$$z-i\bar z=0$$
$$z(1-iz\bar z)=0$$
... | Say $z=re^{i\theta}$ in the polar form.
Putting this value of $z$ in the 2 given equations, we get that
$$z^{8}=|z|^{7} \implies r^{8}e^{i\cdot 8\theta}=r^{7} \implies e^{i\cdot 8\theta}=\frac{1}{r}\tag1$$ $$z=\bar z e^{i\frac{\pi }{2}} \implies z^2=z\bar z e^{i\frac{\pi }{2}} \implies r^{2}e^{i\cdot 2\theta}=r^2e^{i\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If the chord $x+y=b$ of the curve ... If the chord $x+y=b$ of the curve $x^2+y^2-2ax-4a^2=0$ subtends a right angle at the origin, prove that: $b(b-a)=4a^2$
My Approach.
Given,
Equation of the chord,
$$x+y=b$$
$$\frac {x+y}{b}=1$$
Now,
Equation of the curve,
$$x^2+y^2-2ax-4a^2=0$$
$$x^2+y^2-2ax=4a^2$$
$$(b-y)^2+(b-x)^... | First solve the equations simultaneously, and we arrive at the quadratic equation $$2x^2-2x(a+b)+b^2-4a^2=0$$
The roots satisfy $$x_1+x_2=a+b$$ and $$x_1x_2=\frac{b^2-4a^2}{2}$$
The perpendicularity condition can be written as $$x_1x_2+y_1y_2=0$$
$$\implies x_1x_2+(b-x_1)(b-x_2)=0$$
$$\implies 2x_1x_2-b(x_1+x_2)+b^2=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving $\sum_{k=0}^n 2^k \binom{2n-k}{n}=4^n$ using generating functions So I have to show this identity. $$\sum_{k=0}^n 2^k \binom{2n-k}{n}=4^n$$
I tried two techniques to do it. First one would be seeing that it's convolution of two sequences: $\langle 2^n \rangle _{n=0}^\infty$ and $\langle \binom{2n}{n} \rangle _{... | Here is a variation based upon the coefficient of method. We use the symbol $[z^n]$ to denote the coefficient of $z^p$ in a series. This way we can write e.g.
\begin{align*}
[z^p](1+z)^q=\binom{q}{p}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n\binom{2n-k}{n}2^k}&=\sum_{k=0}^{n}[z^n](1+z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How do we integrate $\dfrac{1}{1+x^4}$? I am still confused with this integral: $\int \dfrac{1}{1+x^4}dx$
There one solution was given by a pupil that we write it as $\dfrac{x^2+1-(x^2-1)}{2(1+x^4)}$
Then split into two terms and proceed by putting $x+{1\over x}=t$ and completing the square.
My question is can we appro... | For partial fractions factor into quadratics (this is the hard part):
$$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$$
Do the standard decomposition method with undetermined cofficients:
$$\frac{1}{x^4+1}=\frac{1}{2\sqrt{2}}\left(\frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1}
-\frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1}
\right)$$
Integrate:
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$ If $a,b,c$
are the sides of a triangle and $p,q,r$ are positive real numbers then prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$
After modification. I have to prove $(a^2 p^2+b^2 q^2 + c^2 r^2) \ge pr(a^2+c^2-b^2)+qr(b^2+c^2-a^2)+pq(a^2+b^2-c^2)$
| Let $x=q-r$, $y=r-p$, and $z=p-q$.
Then $x+y+z=0$, and the problem is equivalent to
$$a^2 yz+b^2 zx+ c^2 xy \leq 0.$$
If we substitute $x$ with $-y-z$, then the above inequality is equivalent to
$$(a^2-b^2-c^2)yz \leq b^2 z^2 + c^2 y^2.$$
Because $b-c<a<b+c$,
$$(a^2-b^2-c^2)yz \leq |a^2-b^2-c^2| |yz| < |2bc||yz| \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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What is the maximal value of $2 \sin x - 7 \cos x$? What is the maximal value of $2 \sin x - 7 \cos x$?
How do I calculate this? Do I have to write out the $\sin$ and the $\cos$?
| Hint: one can use formula
$$\sin x \cos y- \cos x \sin y = \sin(x-y),$$
and use such 'tricky' $y_0$, that
$$
\sin y_0 = \dfrac{7}{?}, \;\cos y_0 = \dfrac{2}{?}.
$$
Let $y_0$ is such that $\sin y_0 = \dfrac{7}{\sqrt{7^2+2^2}} = \dfrac{7}{\sqrt{53}}$, and $\cos y_0 = \dfrac{2}{\sqrt{53}}$, namely $y_0 = \arcsin(7/\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find the smallest positive integer that satisfies the system of congruences $N \equiv 2 \pmod{11}, N \equiv 3 \pmod{17}. $ Find the smallest positive integer that satisfies the system of congruences
\begin{align*}
N &\equiv 2 \pmod{11}, \\
N &\equiv 3 \pmod{17}.
\end{align*}
The only way I know to solve this problem is... | By the second constraint $N$ is a number of the form $17k+3$. We may now impose the first constraint:
$$ 17k+3\equiv 6k+3 \equiv 3(2k+1) \equiv 2\pmod{11} $$
leading to $2k+1\equiv 8\pmod{11}$, equivalent to $k\equiv 9\pmod{11}$. It follows that the smallest positive number fulfilling both constraints is given by
$$ \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Defining the foci of "slanted" ellipse equation How to define the foci ($F_1,F_2$) coordinates of the slanted ellipse $x^2+4xy+9y^2=9$?
| $x^2 + 4 x y + 9 y^2 = 9$ or
$\begin{pmatrix}x &y\end{pmatrix}\begin{pmatrix}1&2\\2&9\end{pmatrix}\begin{pmatrix}x \\y\end{pmatrix}=9$
Let's rotate by standard theory. The eigenvalues are $\lambda=5\pm 2\sqrt{5}$ and $P=\begin{pmatrix}\frac1{\sqrt{10+4\sqrt{5}}}&-\frac1{\sqrt{10-4\sqrt{5}}}\\\frac{2+\sqrt{5}}{\sqrt{10+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Expanding $(2^b-1)\cdot(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})$ Can someone please explain to me how the following works (primarily interested in an explanation of the second step when $2^b$ is expanded?
I understand how each series cancels out to equal $2^n-1$ at the end.
$$\begin{align*}
xy&=(2^b-1) \cdot (1 + 2^b + 2^{2... | $\begin{align*}
xy&=(2^b-1) \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
\end{align*}$
Let $A = (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})$ so
$xy = (2^b-1)A = 2^b*A - A$. Let's continue...
$\begin{align*}
&=2^b \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b}) - (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
\end{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Proving the identity $\frac{\cos^2\theta+\tan^2\theta-1}{\sin^2\theta}=\tan^2\theta$ I am stuck with this trigonometric identity. It appeared in a question paper of mine, and I am wondering whether there is a print error or something, because I absolutely have no idea how to solve this.
$$\frac{\cos^2\theta+\tan^2\th... | Follow the usual process of manipulating only one side to make it look like the other side, and leave the other side completely alone. The LHS is much more complicated, so let's start there and try to make it look like the RHS.
\begin{align*}
\frac{\cos^2\theta + \tan^2\theta - 1}{\sin^2\theta}
&= \frac{\cos^2\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate $\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}}$? I need some help with computing this limit:
$$\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}... | What is the problem?
From first inspection:
*
*$\cos(\sin x) + \frac{x^2}{2} \to 1^-$
*$e^{x^2}-1 \to 0^+$
*$1+2x - \sqrt{ (1+2x)^2 - 2x^2 }\to 0^+$
So we have a case of $1^\infty$, which is indeterminate, and therefore we are going to need series expansions to solve this. Let us develop the following exponent t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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In any triangle $ ABC $ $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $. In any triangle $ ABC $ prove that $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $.
Please help. Thanks in advance.
| Note that
\begin{align*}
\frac{\sin(B-C)}{\sin (B+C)}&=\frac{\sin B \cos C-\cos B \sin C}{\sin B \cos C+\cos B \sin C}\\
&=\frac{\frac{\sin B}{\sin C} \cos C-\cos B}{\frac{\sin B}{\sin C} \cos C+\cos B}\\
&=\frac{\frac{b}{c} \frac{c^2-a^2-b^2}{2ab}-\frac{b^2-a^2-c^2}{2ac}}{\frac{b}{c} \frac{c^2-a^2-b^2}{2ab}+\frac{b^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to integrate $ \int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}} $? I am having a little problem with my maths homework. The problem is as follows:
\begin{equation}
\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}
\end{equation}
I tried to do the following but got stuck halfway:
Let $\ \ x \ = asin\theta, \ he... | $$I=\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}$$
Let $x^2+y^2=a^2\implies dx=-\frac{y}{\sqrt{a^2-y^2}}dy$
Thus:
$$I=\int^a_0{\cfrac{dy}{y \ + \ \sqrt{a^2 \ - \ y^2}}}\frac{y}{\sqrt{a^2-y^2}}$$
Adding these together,
$$2I=\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}\left(1+\frac{x}{\sqrt{a^2-x^2}}\right)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed form of $k$-th term in the recurrence $D_j = 1 + z^2\left(1-\frac{1}{D_{j-1}}\right)$, where $D_0 = 1+z^2$ I have $D_0=1+z^2$ and $D_{j}=1+z^2(1-1/D_{j-1})$ for $j>0$. I want to write an expression for arbitrary $D_k$, and what I have is below.
$D_{k}=1+z^2\Bigg(1-\bigg(1+z^2\big(1-(1+z^2(1-(\cdots(1+z^2)^{-1}\c... | To expand upon my comment to @BarryCipra's answer ...
Multiplying numerators and denominators by $(1-z^2)$ gives
$$D_0 = \frac{1 - z^4}{1 - z^2} \qquad D_1 = \frac{1 - z^6}{1-z^4} \qquad D_2 = \frac{1-z^8}{1-z^6}$$
We can show that the pattern continues.
Assuming
$$D_j = \frac{1 - z^{2j+4}}{1-z^{2j+2}} \tag{$\star$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all ordered pairs of real numbers $(x, y)$ such that $x^2y^2 + 2xy^2 + 5x^2 + 3y^2 + 10x + 5 = 0$. How can I find all ordered pairs of real numbers $(x, y)$ such that $x^2y^2 + 2xy^2 + 5x^2 + 3y^2 + 10x + 5 = 0$?
I don't see a pattern, I don't know how to proceed. Any answer is greatly appreciated.
| $$ y^2 \; ( x^2 + 2x + 3) = -5(x^2 + 2 x + 1) $$
$$ y^2 = -5 \; \; \frac{(x+1)^2 }{2 + (x+1)^2} $$
The right hand side is never positive. It is only zero when $x=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{x\to 0}\frac{x}{x+x^2e^{i\frac{1}{x}}} = 1$ $$\lim_{x\to 0}\frac{x}{x+x^2e^{i\frac{1}{x}}} = 1$$
I've trying seeing that:
$$e^{\frac{1}{x^2}i} = 1+\frac{i}{x^2}-\frac{1}{x^42!}-i\frac{1}{x^63!}+\frac{1}{x^84!}+\cdots\implies$$
$$x^2e^{\frac{1}{x^2}i} = x^2+\frac{i}{1}-\frac{1}{x^22!}-i\frac{1}{x^43!}+\frac{1}{x^... | We assume that $x \in \mathbb{R}$. One may write, as $x \to 0$,
$$
\left|\frac{x}{x+x^2e^{i\frac{1}{x}}} \right|=\frac{1}{\left|1+xe^{i\frac{1}{x}} \right|}\le \frac{1}{1-\left|xe^{i\frac{1}{x}} \right|}=\frac{1}{1-\left|x\right|} \to 1
$$ since
$$
||a|-|b||\le |a+b|.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove or disprove $\sum\limits_{cyc}\frac{x^4+y^4}{x+y}\le 3\frac{x^4+y^4+z^4}{x+y+z}$
Let $x,y,z\ge 0$. Prove or disprove
$$\dfrac{x^4+y^4}{x+y}+\dfrac{z^4+y^4}{z+y}+\dfrac{z^4+x^4}{x+z}\le 3\dfrac{x^4+y^4+z^4}{x+y+z}$$
This is what I tried. Without loss of generality, let $x+y+z=1$, then
$$\Longleftrightarrow \su... | Let $x\geq y\geq z$. Hence,
$$y^2\left(\frac{3(x^4+y^4+z^4)}{x+y+z}-\sum_{cyc}\frac{x^4+y^4}{x+y}\right)=y^2\sum\limits_{cyc}\left(\frac{3(x^4+y^4)}{2(x+y+z)}-\sum_{cyc}\frac{x^4+y^4}{x+y}\right)=$$
$$=y^2\sum\limits_{cyc}\frac{(x^4+y^4)(x+y-2z)}{2(x+y+z)(x+y)}=y^2\sum\limits_{cyc}\frac{(x^4+y^4)(y-z-(z-x))}{2(x+y+z)(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Given $abc=1$ and $0< c \leq b \leq1\leq a$, prove that $8(a+b+c)^2\le9(1+a^2)(1+b^2)(1+c^2)$ I can't make progress with proving this inequality.
I have tried opening the brackets and using $abc=1$ in order to obtain the following:
$$a^2+b^2+ \frac 1{a^2b^2}+18+9a^2b^2+\frac 9{a^2}+\frac9{b^2}\ge 16 \left (ab+\frac 1a+... | We begin by observing that if $a=b=c=1$, then the two expressions are equal to each other.
Expand everything. Use the equation $abc=1$ to get rid of compound terms. Gather terms with each variable separately. Use the preceding observation to factor each nominator.
$$\begin{align} & 9(1+a^2)(1+b^2)(1+c^2)-8(a+b+c)^2 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to find the square roots of $z = 5-12i$ I am asked the following question:
Find the square roots of $z = 5-12i$
I know that this problem can be easily solved by doing the following:
$$
z_k^2 = 5-12i\\
(a+bi)^2 = 5-12i\\
(a^2-b^2) + i(2ab) = 5-12i\\
\\
\begin{cases}
a^2 - b^2 = 5\\
2ab = -12
\end{cases} \quad \Rig... | You can apply de Moivre's theorem (second example) on the first example without doubt. Though you need a calculator or a trigonometrical table handy to check the sin and cos of a certain angle which feels kinda painful, at least for me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Easier way to discover the area of a right triangle In the following right triangle: $y-x=5$ and the altitude to the hypotenuse is 12. Calculate its area.
I've managed to discover its area using the following method, but it ends up with a 4th degree equation to solve. Is there an easier way to solve the problem?
$ha=x... | Square both sides of relationship $(x-y)=5$; then apply Pythagoras, giving
$\tag{1}x^2+y^2-2xy=25 \Leftrightarrow a^2-2xy=25$
Besides, the area $S$ of the triangle can be computed in two ways :
$\tag{2}S=\frac{xy}{2}=\frac{12a}{6}=6a$
Plugging the value of $a$ taken from (2) in (1), one gets a quadratic equation in var... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Solve 6th degree polynomial: $(x^2 - 3x - 4)(x^2 - 5x + 6)(x^2 + 2x) + 30$ I came across what seems to be a very difficult "solve for $x$" type of
problem, primarily because there should be $6$ real roots of this problem:
$$(x^2 - 3x - 4)(x^2 - 5x + 6)(x^2 + 2x) + 30 = 0.$$
My first step was to (tediously) expand this ... | Let $P(x)$ denote the LHS. Note that the first summand factors nicely:
$$
P(x)=(x-2)(x-3)(x-4)x(x+1)(x+2)+30.
$$
From this it is clear that $P(x)$ is symmetric around $x=1$. Explicitly, setting $y=x-1$ we have
$$\begin{eqnarray*}
P(x)&=&(y-1)(y-2)(y-3)(y+1)(y+2)(y+3)+30\\
&=&(y^2-1)(y^2-4)(y^2-9)+30.
\end{eqnar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Expressing $\sqrt[3]{7+5\sqrt{2}}$ in the form $x+y\sqrt{2}$ Express $\sqrt[3]{(7+5\sqrt{2})}$ in the form $x+y\sqrt{2}$ with $x$ and $y$ rational numbers.
I.e. Show that it is $1+\sqrt{2}$.
| $$(1+\sqrt{2})^3=(1+2\sqrt{2}+2)(1+\sqrt{2})=(3+2\sqrt{2})(1+\sqrt{2})=3+3\sqrt{2}+2\sqrt{2}+4$$
$$\therefore (1+\sqrt{2})^3=7+5\sqrt{2}$$
$$\Rightarrow 1+\sqrt{2}=\sqrt[3]{7+5\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Closed form solution for infinite summation Thomas, Bruckner & Bruckner, Elementary Real Analysis.
Prove that for all r > 1,
$$\frac{1}{r - 1} = \frac{1}{r+1} + \frac{2}{r^2 + 1} + \frac{4}{r^4 + 1} + \frac{8}{r^8 + 1} + \cdots$$
So far I have $$ \frac{1}{r-1} -\frac{1}{r+1} = \frac{2}{r^2 -1} $$
$$\sum_{n=1}^\infty ... | Notice that
$$\frac 1 {r-1} - \left(\frac 1 {r+1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) = \left( \frac 1 {r-1} - \frac 1 {r+1} \right) - \left(\frac 2 {r^2 + 1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) = \\
\frac 2 {r^2 - 1} - \left(\frac 2 {r^2 + 1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) = \frac 4 {r^4 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
| (x$^2$ - 3x + 1)$^2$ = (2x - 3)$^2$
(x$^2$ - 3x + 1)$^2$ - (2x - 3)$^2$ = 0
(x$^2$-3x+1-2x+3)(x$^2$-3x+1+2x-3) = 0
(x$^2$-5x+4)(x$^2$-x-2) = 0
(x-1)(x-4)(x+1)(x-2) = 0
Therefore, solution is x = 1, x = 4, x = -1 or x = 2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
} |
Find the base numeric system Find the numeric base we are using if $ x = 4 $ and $ x = 7 $ are zeros of $ 5x^2 - 50x + 118. $
When I plug in $ x = 4 $ and $ x = 7, $ I receive $ -2 $ and $ 13, $ respectively, but how do I proceed from that?
| Here's a different way to do it.
$f (x)=5x^2 - 50_bx +118_b =$
$5x (x-b)+118_b $
$f (4)=20 (4-b) +118_b$
$f (7)=35 (7-b)+118_b $
So $118_b= 20 (b-4)=35 (b-7) $
So $15b =7*35-80$
$b=11$.
Or we could note $35|118_b $ and $20|118_b $ so $140|118_b $ and $7|b-4$ and $4|b-7$. If $140 =118_b $ then $b=11$ is a good guess.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Supremum of rounded function Can you please help me to find and prove $\sup\{\sqrt{n}- \left\lfloor \sqrt{n}\right\rfloor : n\in N\}$?
I assume that it is something like $1$ or some other constant, but not quiet sure.
| To prove that $$\sup \left\{ \sqrt{n} - \left\lfloor \sqrt{n} \right \rfloor : n \in \Bbb{N} \right\} = 1$$
(note that I use the more modern symbol $\lfloor x \rfloor$ in place of $[x]$ to mean the highest integer not exceeding $x$)
Step 1: For $n \in \Bbb{N}$ or indeed any real value of $n$, $\sqrt{n} - \left\lfloo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $f(x+1) = x^2 + 3x +5$, then find $f(x)$ A challenge problem from precalc class. I don't know what to do with the one from $f(x+1)$, it doesn't factor well, I could do completing the square but then how do I find $f(x)$? Just stuck on this.
| Just to be different.
$$f(x+1) = x^2 + 3x + 5$$
\begin{align}
f(x) - f(x+1)
&= f((x-1)+1) - f(x) \\
&= (x-1)^2 + 3(x-1) + 5) - (x^2 + 3x + 5) \\
&= ((x-1)^2 - x^2) + (3(x-1) - 3x) + (5-5) \\
&= (x-1-x)(x-1+x) - 3\\
&= (-1)(2x-1) - 3 \\
&= -2x - 2
\end{align}
So $f(x) = f(x+1) - 2x - 2 = x^2 + x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1951368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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System of equations involving square roots Hallo :) I am hopeless with this exercise:
Solve the system of equations over the positive real numbers
$$\sqrt{xy}+\sqrt{xz}-x=a$$
$$\sqrt{zy}+\sqrt{xy}-y=b$$
$$\sqrt{xz}+\sqrt{yz}-z=c$$
where $a, b, c$ are positive real numbers.
I tried to +,- and / the equations with one ... | We can solve the system in the following way (though I'm not sure if it is "reasonable") :
We have
$$\sqrt y+\sqrt z-\sqrt x=\frac{a}{\sqrt x}\tag1$$
$$\sqrt z+\sqrt x-\sqrt y=\frac{b}{\sqrt y}\tag2$$
$$\sqrt x+\sqrt y-\sqrt z=\frac{c}{\sqrt z}\tag3$$
From $(1)$,
$$\sqrt z=\sqrt x-\sqrt y+\frac{a}{\sqrt x}\tag4$$
From ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Necessary conditions for a Sudoku puzzle to have no repetitions Is it true that if a Sudoku puzzle has the following features there will be no repetitions in rows, columns and $3 \times 3$ subsquares?
*
*The sum of each row must be $45$
*The sum of each column must be $45$
*The sum of each $3 \times 3$ subsquare m... | No. For instance, this "sudoku" fulfills your conditions, but has some repetitions:
$$
\begin{array}{|ccc|ccc|ccc|}
\hline 5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5\\
\hline 5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5\\
\hline 5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Lemma on switching between mod $p$ and mod $p^2$ or mod $p^3$ Can someone help me prove the following lemma? Also can it be strengthened?
Let $p\geq 5$ be a prime number. Prove that if $p|a^2+ab+b^2$, then $p^3|(a+b)^p-a^p-b^p$
Here is what I tried:
We want to show $$p^3|a^{p-1}\binom{p}{1}b+a^{p-2}\binom{p}{2}b^2+\dot... | Here is the proof anticipated by Stefan4024, based on this question linked to in his answer.
We first show that if $p \equiv 1 \pmod 6$, then
$$
p(a^2 + ab + b^2)^2 \,\mid\, (a+b)^p - a^p - b^p .
$$
Consider some fixed $b$, and let $f(x)$ be the polynomial
$$
f(x) = (x + b)^p - x^p - b^p.
$$
Let $\omega = e^{2\pi i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1961526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Prove there exists a point O such that $ OB \leq \frac {2}{\sqrt {3}} BY, OD \leq \frac {2}{\sqrt {3}} DZ, OF \leq \frac {2}{\sqrt {3}} FX$
Let the incircle of BDF touch DF, FB, BD at X, Y, Z respectively. Prove there exists a point O such that $$ OB \leq \frac {2}{\sqrt {3}} BY, OD \leq \frac {2}{\sqrt {3}} DZ, OF ... | I'm a big fan of the tangent half-angle formula. Without loss of generality, the incircle is the unit circle, rotated such that none of $X,Y,Z$ lies at $(-1,0)$. Then one may choose coordinates as
\begin{align*}
X&=\frac1{1+x^2}\begin{pmatrix}1-x^2\\2x\end{pmatrix}&
Y&=\frac1{1+y^2}\begin{pmatrix}1-y^2\\2y\end{pmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1961716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove the limit of a sequence by definition: $\lim((n^2 + 1)^{1/8} - n^{1/4}) = 0$? I have to prove the limit of a sequence using $\epsilon - N$ definition: $$\lim_{n\to\infty}((n^2 + 1)^{1/8} - n^{1/4}) = 0$$.
Attempt:
We want to show: $\forall \epsilon>0$ $\exists N \in \mathbb{N}$, s.t. if $n \ge N$, then $|... | Our expression equals $(n^2+1)^{1/8} - (n^2)^{1/8}.$ By the mean value theorem,
$$\tag 1 (n^2+1)^{1/8} - (n^2)^{1/8} =(1/8)c_n^{-7/8}\cdot 1,$$
where $c_n\in (n^2, n^2+1).$ It follows that $(1) \le (1/8)(n^2)^{-7/8} = (1/8)n^{-7/4}.$ You're now set up for an $\epsilon$-$N$ proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Change of polynomial basis exercise Can someone please check if my solution is correct? Thanks.
Considering the following basis:
$$
B= \left\{ x^2,1+x^2,x-1 \right\}\\
C= \left\{ x,1-x,x^2 \right\}
$$
a) Find the change of basis matrices from $B$ to $C$ and $C$ to $B$.
b) Find the coordinates of the vector $x+1$ in $C$... | Everything looks correct. The only thing I would say is that you should make sure D is in fact a basis by checking the vectors are linearly independent and checking whether they in fact span $R^2[x]$. This should be fairly easy by looking at the dimension of the subspace or the linear independence of the coordinates.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac... |
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
There is a solution if and only if one of the two equations $x^2-ax+1=\pm\dfrac14$ has two roots $r,s$ and $|r-s|\ge 2$ and the solution of $f(x+2)=f(x)$ satisfy $\left|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 10,
"answer_id": 9
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Minimal value of $\sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$ without derivatives and without distance formula Let $f(x) = \sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$, where all coefficients are real.
It can be shown using a distance formula that the minimal value of $f(x)$ is
$D = \sqrt{(a-c)^2+(|b|+|d|)^2}$.
Show ... | Applying Fermat's Principle on reflection:
Assume $a<c$,
\begin{align*}
T(x) &= \sqrt{(x-a)^2+b^2}+\sqrt{(x-c)^2+d^2} \\
T'(x) &=\frac{x-a}{\sqrt{(x-a)^2+b^2}}-\frac{c-x}{\sqrt{(x-c)^2+d^2}} \\
0 &= \sin i-\sin r \\
i &= r \\
\end{align*}
If you've accepted the law of reflection, then
\begin{align*}
\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trying to find shortest distance from point $(2,0-3)$ to $x+y+z=1$ I'm trying to find shortest distance from the point $(2,0-3)$ to $x+y+z=1$.
I found $d^2=(x-2)^2+y^2+(z+3)^2$
Substituting for $z$: $d^2=(x-2)^2+y^2+(-x-y+1+3)^2$
$f_x=2x-4-2(-x-y+4)$ and $f_y=2y-2(-x-y+4)$ Next would be finding zeros as the critical p... | We have $4x+2y=12$ and $2x+4y=8$. First equation reduces to $y=6-2x$. Plugging that into the second gives $2x+4(6-2x)=8$ which solves to $2x+24-8x=8$ or $x=8/3$. Plugging this answer into $y=6-2x$ yields $y=2/3$. This is one of the many ways to solve such a system. Anyway, with this info you can find $z$ and from here ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that the chord of the ellipse passes through a fixed point Variable pairs of chords at right angles are drawn through a point $P$ (forming an angle of $\pi/4$ with the major axis) on the ellipse $\frac {x^2}{4}+y^2=1$, to meet the ellipse at two points $A $ and $B $. Prove that the line joining these two points p... | This is a property of general ellipses, so let's consider the ellipse with major and minor radii $a$ and $b$. For a given point $P = (a \cos 2\theta, b \sin 2\theta)$, we'll identify the point $Q$ common to all chords $\overline{AB}$ such that $\overline{AP}\perp\overline{BP}$.
It's straightforward to find the coordin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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How to find results for $-z + \frac {3} {\overline {z}}=2$ Can you just please help me solve this problem, because i don't know how to solve it: $$-z + \frac {3} {\overline z}=2$$
| $$-z + \frac {3} {\overline z}=2$$
$$-z \overline z+3=2\overline z$$
As $z \overline{z}=\left|{z^2}\right|$,
$$
\begin{align}
-\left|{z^2}\right|+3&=2 \overline{z}\\
&= 2a-2bi \\
\end{align}
$$
Equating imaginary parts,
$$2\Im{(z)}=0\implies 2b=0\implies b=0$$
Equating real parts,
$$
\begin{align}
2 \Re{(z)} &= -(a^2+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
If two consecutive numbers are removed from the series $1+2+3+\ldots+n$ the average becomes $99/4$. Find the two numbers. The initial average will be $\frac{n+1}{2}$. If the two numbers are $k$ and $k+1$ then the new average will be $\frac{n(n+1)/2-(2k+1)}{n-2}$. I couldn't figure further even though I got the relation... | We have, that the sum of $n+1$ terms, excluding the $m$-th and $m+1$-th, is:
$$
\begin{gathered}
S(n + 1,m) = \sum\limits_{1\, \leqslant \,k\, \leqslant \,m - 1} k + \sum\limits_{m + 2\, \leqslant \,k\, \leqslant \,n + 1} k = \sum\limits_{1\, \leqslant \,k\, \leqslant \,m - 1} k + \sum\limits_{1\, \leqslant \,k\, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Proof of $\sum_\limits{n = 1}^{\infty} {k\choose n}{b-1 \choose n-1}x^n \sim \frac{x^b k^b}{\Gamma(1+b)} \quad \text{as}~k \to \infty$ I want to prove the following aymptotic result:
$$\sum_{n = 1}^{\infty} {k\choose n}{b-1 \choose n-1}x^n \sim \frac{x^b k^b}{\Gamma(1+b)} \quad \text{as}~k \to \infty,$$
where $ k \in \... | Define
$$
f_k(x)=\sum_{n=1}^\infty\binom{k}{k-n}\binom{b-1}{n-1}x^{n-1}\tag{1}
$$
then we are looking for the asymptotic expansion of $x\,f_k(x)$.
Vandermonde's Identity and Gautschi's Inequality say
$$
\begin{align}
f_k(1)
&=\sum_{n=1}^\infty\binom{k}{k-n}\binom{b-1}{n-1}\\
&=\binom{k+b-1}{k-1}\\
&=\frac{\Gamma(k+b)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find $24 \cot^2 x$
Suppose $x$ is in the interval $[0,\pi/2]$ and $\log_{24 \sin x} (24 \cos x) = \frac{3}{2}$. Find $24 \cot^2 x$.
From the given equation we have $24 \cos{x} = (24 \sin{x})^{\frac{3}{2}}$ and so $24\cot^2{x} = 24^2\sin{x}$. How do we continue?
| You have $(24\cos x)^2 =(24\sin x)^3,\;$ or $\cos^2 x =24\sin^3 x.\;$
With $y=\sin x$ this gives the cubic $24y^3=1-y^2$ with the real solution
$y_1 = \frac{1}{3} \;$ or
$$x=\arcsin\left(\frac{1}{3}\right) \approx 0.3398369$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $10^{3n+1}+10^{6n+2}+1=111k$ I'm stuck to prove this one can anyone help me please?
Prove that $10^{3n+1}+10^{6n+2}+1=111k$.
I'm not sure what exactly I should do.
n is a natural number and k is an integer
| The OP has corrected the problem. The corrected answer is below.
Assuming that $k$ must be an integer (not stated by OP), this is the same as showing that
$$
10^{n+1}+10^{6n+2}+1\equiv 0\pmod{111}.
$$
Observe that $10^3=1000=999+1=1+9(111)$. Therefore, there are only three values for the first and second terms, $10$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Modular arithmetic three variables Show that if the integers $x, y,$ and $z$ satisfy $x^3 + 3y^3 = 9z^3$
then $x = y = z = 0.$
How should I interpret this question and how to proceed?
I am thinking about the Euclidean algorithm but it becomes confusing when $x,y,z$ comes like variables?
| First notice that if $d=\mbox{gcd}(x,y,z)$ then $d^3$ can be factored out of the equation. So we can assume that $d=1$. Then $x^3 = 9z^3-3y^3$, so $3$ divides $x$, say $ x=3k$. So we have $3^3k^3 = 9z^3-3y^3$ and we can divide everything by $3$ to get $9k^3= 3 z^3-y^3$. A similar argument shows $3$ divides $y$. R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Matrix with unknown coefficient $a$ - determine number of solutions for each $a$ I have a matrix that looks like this:
\begin{array}{cccc|c}
a & 1 & 0 & 0 & 1 \\
1 & a & 0 & 0 & 2 \\
0 & 0 & a & 2 & 1 \\
0 & 0 & 2 & a & 1
\end{array}
I then proceeded to row reduce the matrix so it looks like this:
\begin{array}{cccc|c}... | Swap rows 1 and 2, and rows 2 and 4. You obtain an augmented matrix for which it's easier to apply pivot's method:
\begin{align}
\begin{bmatrix}
1&a&0&0&2\\a&1&0&0&1\\0&0&2&a&1\\0&0&a&2&1
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&a&0&0&2\\0&1-a^2&0&0&1-2a\\0&0&2&a&1\\0&0&0&4-a^2&2-a
\end{bmatrix}
\end{align}
Thus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it inequality true Please help me prove the inequality:
$$\sqrt[3]{3+\sqrt[3] 3}+\sqrt[3]{3-\sqrt[3] 3}<2\sqrt[3] 3$$
Thanky for your help and your attention.
| We can use $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$.
Since $a^2+b^2+c^2-ab-ac-bc=0\Leftrightarrow a=b=c$
and $\sqrt[3]{3+\sqrt[3]3}\neq\sqrt[3]{3-\sqrt[3]3}$ , we need to prove that
$$3+\sqrt[3]3+3-\sqrt[3]3-8\cdot3+3\cdot2\sqrt[3]3\sqrt[3]{9-\sqrt[3]9}<0$$ or
$$\sqrt[3]{27-3\sqrt[3]9}<3$$
which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square. The famous IMO problem 6 states that if $a,b$ are positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $\frac{ a^2 + b^2}{ab + 1 }$ is a perfect square, namely, $gcd(a,b)^2$. How about a ... | Here is a bit of a hacky answer to your question in the affirmative
You observe that $k = \mathrm{gcd}(a,b)^2$. After plodding through various resources, it is because one can Viete Jump:
$$ (a,b) \mapsto \big(a_1, b_1\big) \mapsto \dots \mapsto (k,0) $$
and the gcd is conserved. Once I agree with you, let $a = \tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 1,
"answer_id": 0
} |
Solve Recurrence Relation for Maximum and Minimum in an Array. I know that recursive equation of this algo is
$T\left ( n \right )=2T\left ( \frac{n}{2} \right )+2 $
Given that $T\left ( 1 \right )=0,T\left ( 2 \right )=1 $
and its solution is also given here, ijust want to clear my doubt where i am stuck at..
I solve... | You can check as followed that your solution fits the recurrence like this:
$T(1)=2\cdot 1-2=0$
$T(n) = 2n-2 = 2(2\frac{n}{2}-2) +2 = 2\cdot T(\frac{n}{2}) + 2$
Similarly you can check that the answer you have been given fits:
$T(2)=1$
$T(n) = 2\cdot T(\frac{n}{2}) + 2$
How to reconcile these 2 facts? Well the 2 base-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How prove this $BC$ always passes through a fixed point with $\frac{x^2}{4}+y^2=1$
if the point $A(0,1)$ on the ellipse $\Gamma:$ $\dfrac{x^2}{4}+y^2=1$ and the circle $\tau:$ $(x+1)^2+y^2=r^2(0<r<1)$,if $AB,AC$ tangent the circle $\tau$ ,$B,C\in \Gamma$,show that the line $BC$ always passes through a fixed point
I... | The equation of the line passing through $A(0,1)$ is given by $mx-y+1=0$.
Since we want this line to be tangent to the circle, we have
$$r=\frac{|m(-1)-0+1|}{\sqrt{m^2+(-1)^2}},$$
i.e.
$$(r^2-1)m^2+2m+r^2-1=0\implies m_1+m_2=\frac{2}{1-r^2},\quad m_1m_2=1\tag1$$
Eliminating $y$ from $mx-y+1=0$ and $x^2/4+y^2=1$ gives
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rig... | The best way to show this kind of an equality, is to show that $S_{k-1} - S_k$ is divisible by $9$.
\begin{split} S_{k+1} - S_k & = (5 \times 10^{k+1} + 10^{k} + 3) - (5 \times 10^k + 10^{k-1} + 3)\\ & = 5(10^{k+1}-10^k) + (10^k-10^{k-1}) \\ & = 5\times 9 \times 10^k + 9 \times 10^{k-1} \\ & = 9 \times (5 \times 10^k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Right Triangle: Hypotenuse and Side differ by 1 So I have search to the best of my abilities but cannot find mathematically why this is true and if it is called something specific, the closest thing would be Pythagorean Triples but this is not the case although a (3,4,5) is a triple. I will do my best to explain and ap... | For all $a,b$
$$b^2 - a^2 = (b+a)(b-a)$$
Assuming as you do that $b=a+1$
$$b^2 - a^2 = (b+a)(b-a) = (2a+1)(1)=2a+1$$
and
$$a+b=a+a+1=2a+1$$
so yes, when $b=a+1$, $b^2 - a^2 = a + b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Determinant with variables What is the following determinant?
$$\begin{vmatrix}1+a & b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}$$
I calculated it as $0$ but I do not think it is right. Thanks in advance.
| $$\begin{vmatrix}1+a& b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}=\begin{vmatrix}1 & b & c & d \\0 & 1+b & c & d \\0 & b & 1+c & d \\0 & b & c & 1+d \end{vmatrix}+\begin{vmatrix}a & b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}\\
=\begin{vmatrix}1+b & c ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$? If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?
All I know is that $\sin^{3}a+\cos^{3}a$ is equal to
$$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \... | ${x^3+y^3 \over x+y} = x^2+y^2-xy$.
$(x+y)^2 = x^2+y^2+ 2 xy$, and so
${x^3+y^3 \over x+y} = x^2+y^2 -{1 \over 2} ((x+y)^2-(x^2+y^2))$, from which we get
$x^3+y^3 = (x+y) (x^2+y^2 -{1 \over 2} ((x+y)^2-(x^2+y^2)))$.
Since $x+y = 1.2, x^2+y^2 = 1$ we get $x^3+y^3 = (1.2)(1-{1 \over 2}((1.2^2-1)) = {117 \over 125}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
... | Using Fermat's little theorem, $$17^{40}\equiv 1\pmod{41}\Rightarrow 17^{39}\cdot17\equiv 1\pmod{41}$$ Hence $17^{39}$ is the searched inverse. We have $$17^2\equiv 2\pmod{41}\Rightarrow
17^{30}\equiv 2^{15}\equiv 9\pmod{41}$$
$$17^9\equiv(17^2)^4\cdot 17\equiv 16\cdot17\equiv26\pmod{41}$$Hence $$17^{39}\equiv9\cdot26... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 7
} |
How to calculate width and height of a 45° rotated ellipse bounded by a square? I'm coming from a programming background so I apologies if this is blindingly simple or I misuse terms. I have an ellipse bounded by a square. For simplicity the centre of the square and ellipse is the origin (0,0) while the square is 2 wid... | Consider an square formed by the sides of $\pm x\pm y=m$ (i.e. side length $m\sqrt2$).
Now find an ellipse which touches the square. Consider the top right quadrant. Edge of square is $$L: \quad x+y=m$$Assume point of tangency is $P(h,k)$.
Ellipse, $E$:
$$\begin{align}
\frac {x^2}{a^2}+\frac {y^2}{b^2}&=1\\
\frac {dy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How can I solve $T(n) = \lfloor \frac {n} {2}\rfloor + T(\lceil \frac{n}{2} \rceil)$? I came u with the following equation:
$T(2) = 1$
$T(n) = \lfloor \frac {n} {2}\rfloor + T(\lceil \frac{n}{2} \rceil) | n ∈ \mathbb N$
But I can't find a way to solve it. Is there one?
I need to proof that the solution is right, so I ... | You can solve this as follows. First calculate the first few values
\begin{align}
T(3) = 1 + T(2) = 2 \\
T(4) = 2 + T(2) = 3 \\
T(5) = 2 + T(3) = 4 \\
T(6) = 3 + T(3) = 5
\end{align}
Ok, it might be that T(n) = n - 1. Let's try to prove this by induction. Firstly, we have $T(2) = 1$ so this is obviously true for $n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Unique pair of positive integers $(p,n)$ satisfying $p^3-p=n^7-n^3$ where $p$ is prime
Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.
Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\te... | Given that $p,n\in\mathbb{Z}^+$, where $p$ is a prime, such that $p^3-p=n^7-n^3$.
This implies that $p(p^2-1)=n^3(n^2-1)(n^2+1)\implies p|n^3$ or $p|n^2-1$ or $p|n^2+1$.
Case 1: $p|n^3$. This implies that $p|n\implies n=pk$ for some positive integer $k$. Therefore we have $n^7=p^7k^7$ and $n^3=p^3k^3$. Therefore $$p^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
convergence of $\sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{ \sinh \frac{1}{n}}{\sin\frac{1}{n}} \right )$ $ \sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{\sinh\frac{1}{n}}{\sin\frac{1}{n}} \right ) $
Find the value of $\alpha$ for which the series converges
please, I have no idea to approach the solution... | Hint: use Taylor series, again, again, and again.
In detail:
From a Taylor series around $0$, since $\frac{1}{n}\xrightarrow[n\to\infty]{}0$:
$$\sinh \frac{1}{n} = \frac{1}{n} + \frac{1}{6n^3} + o\left(\frac{1}{n^3}\right) \tag{1}$$
and
$$\sin \frac{1}{n} = \frac{1}{n} - \frac{1}{6n^3} + o\left(\frac{1}{n^3}\right)\tag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $a_{ij}=\max(i,j)$, calculate the determinant of $A$
If $A$ is an $n \times n$ real matrix and
$$a_{ij}=\max(i,j)$$
for $i,j = 1,2,\dots,n$, calculate the determinant of $A$.
So, we know that
$$A=\begin{pmatrix}
1 & 2 & 3 & \dots & n\\
2 & 2 & 3 & \dots & n\\
3 & 3 & 3 & \dots & n\\
\vdots & \vdots & \vdots ... | Apply row operations $R_{j} \leftarrow R_{j} - R_{1}, \, \forall j \in \left \{2, ...n\right \}$ on $A$ to get the following matrix:
$A_1 = \begin{pmatrix}
1& 2 & 3 & ... & n-1 & n\\
1& 0& 0 & ... & 0& 0\\
2& 1& 0 & ... & 0& 0\\
...& ...& ... & ... & ...& ...\\
n-2& n-3 & n-4 & ...& 0& 0\\
n-1& n-2&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
} |
Probability of no complete losses and no complete wins In the NFL, a division consists of 4 teams, each of which plays each other team twice.
Assume that in any game, either team is equally likely to win (and there are no ties). What is the probability that, at the end of the season, the division has neither a perfect ... | You need to consider the important outcomes and their probabilities
*
*The probability that Team A wins all $6$ matches is $\dfrac{1}{2^6}$
*The probability that Team D loses all $6$ matches is $\dfrac{1}{2^6}$
*The probability that Team A wins all $6$ matches and Team D loses all $6$ is $\dfrac{1}{2^{10}}$
*The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2014592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Knowing a quartic has a double root, how to find it? I have a depressed quartic polynomial with three free parameters in the real numbers:
$x^{4}+qx^{2}+rx+s$
Furthermore, the discriminant is constrained to be zero and there are four real roots, exactly two of which are equal. I am only interested in the double root.
... | Yes, there is. You know that your roots (including multiplicities) can be expressed as $\{a, a, -a+δ, -a-δ\}$ (because the sum is zero after the normalization). Expand
$$(x-a)^2(x+a-δ)(x+a+δ) = x^4 - (2a^2 + δ^2)x^2 + 2aδ^2 x + a^4-δ^2a^2$$
and compare with your form. Next, note that the equations
$$\begin{aligned}
- (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\sum _{n=1}^{\infty } \sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$ How can we evaluate:
$$\sum _{n=1}^{\infty } \sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$$
I tried to transform it into $$\sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$... |
Note that $$\sinh^{-1} x-\sinh^{-1} y=
\sinh^{-1} \left( x\sqrt{1+y^2}-y\sqrt{1+x^2} \right)$$
\begin{align*}
\frac{1}{\sqrt{2^{n+2}+2}+\sqrt{2^{n+1}+2}} &=
\frac{\sqrt{2^{n+2}+2}-\sqrt{2^{n+1}+2}}
{(2^{n+2}+2)-(2^{n+1}+2)} \\
&= \frac{\sqrt{2^{n+2}+2}-\sqrt{2^{n+1}+2}}
{2^{n+1}} \\
&= \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Problem Statement:-
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Attempt at solution:-
Let $\alpha=\sqrt{3x^2-7x-30}\;\;\; \text{&} \;\;\;\beta=\sqrt{2x^2-7x-5}$.
Then, we have $$\alpha-\beta=x-5\tag{1}$$
And, we have $$\alpha^2-\beta^2=x^2-2... | As $(\sqrt{x^2-7x-30})^2-(\sqrt{2x^2-7x-5})^2=(x+5)(x-5),$
$$\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$$
$$\implies \sqrt{3x^2-7x-30}+\sqrt{2x^2-7x-5}=x+5$$
Adding we get $$x=\sqrt{3x^2-7x-30}\ \ \ \ (1)$$
Squaring $(1)$ we get $$0=2x^2-7x-30=(2x+5)(x-6)$$
Now for real $x,x=\sqrt{3x^2-7x-30}\ge0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ?? I'd like to evaluate the series $$1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$$
Since I am a high school student, I only know how to p... | Can you prove that $(n-3)(n-2)(n-1)(n) = \frac{(n-3)(n-2)(n-1)(n)(n+1)}{5} - \frac{(n-4)(n-3)(n-2)(n-1)(n)}{5}?$
From there you can substitute different values of $n$ and derive a formula for your expression, and the final summation will be given by
$$ \frac{(n-3)(n-2)(n-1)(n)(n+1)}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
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How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1... | $$\cos^6x+\sin^6x=(\cos^2x)^3+(\sin^2x)^3=$$
$$=(\cos^2x+\sin^2x)(\cos^4x-\cos^2x\sin^2x+\sin^4x)=$$
$$=\cos^4x-\cos^2x\sin^2x+\sin^4x=$$
$$=((\cos^2x)^2+(\sin^2x)^2+2\cos^2x\sin^2x)-3\cos^2x\sin^2x=$$
$$=(\cos^2x+\sin^2x)^2-3\cos^2x\sin^2x=$$
$$=1-3\cos^2x\sin^2x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 2
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Prove that if $a$ is odd, then $a^2\equiv 1\pmod 8$ Prove that if $a$ is odd, then $a^2\equiv 1\pmod 8$I got this question in my discrete mathematics class, can anyone help me?Thanks.
| The clever way to do it is:
$a$ odd $\implies a -1$ and $a+1$ are two consecutive even numbers. As they are consecutive one of them is divisible by $4$ and the other is divisible by $2$. We have no way of knowing which is which but it doesn't matter. One is divisible by $2$ and the other is divisible by $4$ and mult... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Find all $n\in\mathbb N$ such that $\sqrt{n^2+8n-5}$ is an integer. $n^2+8n-5$ has to be a perfect square.
How to find all $n$?
| $y^2=n^2+8n-5 = (n+4)^2 - 21$ implies $21 = x^2-y^2$, for $x=n+4$.
Write $21 = x^2-y^2= (x-y)(x+y)$. Since $x=n+4 \ge 4$, we must have $x+y=7$ or $x+y=21$.
*
*$x+y=7$ implies $x-y=3$ and so $x=5$ and $n=1$
*$x+y=21$ implies $x-y=1$ and so $x=11$ and $n=7$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $N = \frac{b^n-1}{b-1}$ is a pseudoprime number to the base b. Let $n$ a pseudoprime-number to the base $b$ and $\gcd(b - 1, n) = 1$
Prove that $N = \frac{b^n-1}{b-1}$ is a pseudoprime number to the base b.
My Atachment:
Proof. Note that $Φ_p(b) = M_p (b)$ and $\gcd(p, Φ_p(b)) = 1$.
So $P_p(b) = M_p(b)$
Let ... | I have an alternative approach.
We want to see that:
\begin{equation*}
b^{N-1}\equiv 1 ( \text{ mod } N)
\end{equation*}
lets begin by proving that $n|N-1$.
$n$ is pseudoprime, i.e
$$b^{n-1} \equiv 1 ( \text{ mod } n)$$
Therefore $n|( b^{n-1} -1 )$ which is equivalent to say that $n| b^n-b$.
On the other hand, we k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}$, then calculate $14S$.
If $$S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}\,$$ find the value of $14S$.
The question can be simplified to:
Find $S=\sum\limits_{k=1}^n\,t_k$ if $t_n=\dfrac{n}{1+n^2+n^4}$.
As $1+n^2+n^4$ f... | From the first four terms, you can easily guess the pattern: The numerators are $1,3,6,10$ which look like $n(n+1)/2$ and the denominators are $3,7,13,21$ which look to have a constant second difference of $2$ and are thus probably $n^2+n+1$.
This motivates trying to prove, by induction, that
$$
S_n = \frac{n(n+1)}{2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Conjecture for the value of $\int_0^1 \frac{1}{1+x^{p}}dx$ While browsing the post Is there any integral for the golden ratio $\phi$?, I came across this nice answer,
$$ \int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\pi\,\phi}5$$
it seems the general form is just
$$p \int_0^\infty \frac{1}{1+x^{p}}dx=\color{blue}{\frac{\pi... | I'm only going to address Question 1.
The expression proposed in Question 1 is true. However, it is a little bit too complicated than necessary. A simpler version of the expression is
$$2p\int_0^1 \frac{dx}{1+x^p} = \psi\left(\frac{1}{2p} + \frac12\right) - \psi\left(\frac{1}{2p}\right)$$
From reflection formula, tak... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Prove:$x^2+2xy+3y^2-6x-2y\ge-11\;\;\forall(x,y)\in\Bbb{R}$
Problem Statement:-
Prove that for all real values of $x$ and $y$
$$x^2+2xy+3y^2-6x-2y\ge-11$$
I have no idea how to approach this question all I could think on seeing it was tryin to find the linear factors, turns out that the determinant
$$\begin{vmatrix... | We have
$$\begin{align}x^2+2xy+3y^2-6x-2y+11&=x^2+(2y-6)x+3y^2-2y+11\\\\&=(x+y-3)^2-(y-3)^2+3y^2-2y+11\\\\&=(x+y-3)^2+2y^2+4y+2\\\\&=(x+y-3)^2+2(y+1)^2\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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(CHMMC #8 Individual, 2016) Find the smallest $n$ such that $n^2 \,\% \, 5 < n^2 \,\%\, 7 < n^2 \,\%\, 11 < n^2 \,\%\, 13?$
Define $n\,\%\,d$ as the remainder when n is divided by d. What is the smallest
positive integer n, not divisible by 5, 7, 11, or 13, for which $n^2 \,\%
\, 5 < n^2 \,\%\, 7 < n^2 \,\%\, 11 < n... | I found an elementary solution only using congruences and trial-and-error. Given $x\in\Bbb{N}$ such that $5\not\mid x$ and $7\not\mid x$, then $x^2\equiv 1,4\pmod 5$ and $x^2\equiv 1,2,4\pmod 7$. So if $n^2\equiv 4\pmod 5$ it's impossible to have $n^2 \,\%
\, 5 < n^2 \,\%\, 7$. Now, $n^2\equiv 4\pmod 5$ iff $n\equiv2, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all solutions to $x^2\equiv 1\pmod {91},\ 91 = 7\cdot 13$ I split this into $x^2\equiv 1\pmod {7}$ and $x^2\equiv 1\pmod {13}$.
For $x^2\equiv 1\pmod {7}$, i did:
$$ (\pm1 )^2\equiv 1\pmod{7}$$ $$(\pm2 )^2\equiv 4\pmod{7}$$ $$(\pm3 )^2\equiv 2\pmod{7}$$ Which shows that the solutions to $x^2\equiv 1\pmod {7}$ are ... | Hint: Consider the possibility that $x \equiv 1 \pmod 7$ but $x \equiv -1 \pmod {13}$, and so on. (In other words, your mistake was to assume that the $\pm1$ modulo $7$ was the same sign as $\pm1$ modulo $13$.)
Also note that for any prime $p$, if $x^2 \equiv 1 \pmod p$, then we can rewrite this as $$x^2 - 1 \equiv (x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $ Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $
$$\tan \theta +\sec \theta =1.5 $$
$$2\tan \theta +2\sec \theta =3 $$
$$2\sec \theta =3-2\tan \theta$$
$$4\sec^2 \theta =(3-2\tan \theta)^2$$
$$4+4\tan^2 \theta =9-12\tan \theta+4\tan^2 \theta$$
So I get $$\tan... | No one would have any doubt in first solution. In the second solution we have no problem with $\sin\theta=5/13$, but $\sin\theta=-1$ gives $cos\theta=0$ and when you apply $cos\theta=0$ in the expression $\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta} =\frac{3}{2}$ you will find that you have done a mistake bec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Trouble with a substitution I'm struggling to show that
$$ \int_{-1}^{1} \frac{(1-x^2)^{1/2}}{1+x^2} dx $$
to
$$ -\pi + \int_{-\pi}^{\pi} (1+\cos^2(\theta))^{-1}d\theta$$
with $x=\cos(\theta)$
I'm aware I'm missing something obvious but I end up with a stray $\sin(\theta)$
| After the usual substitution, you get
$$
\int \frac{\sqrt{1-x^2}}{1+x^2}dx = \int_0^\pi \frac{\sin^2\theta}{1+\cos^2 \theta}d\theta = \int_0^\pi \frac{1-\cos^2\theta}{1+\cos^2 \theta}d\theta = \int_0^\pi -1 + \frac{2}{1+\cos^2\theta} = \\-\pi + \int_{-\pi}^\pi \frac{1}{1+\cos^2\theta}d\theta.
$$
In the last steps I wa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
| We have $5\over{x+3}$ + $2\over{2x+6}$ = 4.
This can be rewritten as $10\over{2x+6}$ + $2\over{2x+6}$ = 4.
Multiplying by 2x + 6 on each side, we get: 10 + 2 = 4(2x + 6)
Dividing by 4 on each side: 3 = 2x + 6
$\Rightarrow$ -3 = 2x $\Rightarrow$ x = $-3\over{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
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the sum of consecutive odd numbers If the sum of consecutive odd numbers starting with $-3$ until $2k+1$ equals $21$
What is the value of $k$ ?
I can solve this by trying the numbers $-3-1+1+3+5+7+9=21$ , so the last term is $7th$ so the $k$ value is $3$
But I could not solve this with formula, I know the odd numbers c... | You can reduce this sequence: $-3-1+1+3+5+7+9=21$.
$-3-1+1+3=0$. So you are being left with:
$5+7+9=21$. Since the last term is $9$ and the last term in your question is $2k+1$ we have:
$2k+1=9$
$2k=9-1$
$k=8:2$
$k=4$
Edit:
I would like to add, that your confusion might come from how you are understand "until $2k+1$". ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\frac1{a+b+1}+\frac1{b+c+1}+\frac1{c+a+1}\le1$ If $abc=1$ then
$$\frac1{a+b+1}+\frac1{b+c+1}+\frac1{c+a+1}\le1$$
I have tried AM-GM and C-S and can't seem to find a solution. What is the best way to prove it?
| Also we can use the following reasoning.
For positives $a$, $b$ and $c$ let $a=x^3$, $b=y^3$ and $c=z^3$.
Hence, $\sum\limits_{cyc}\frac{1}{a+b+1}=\sum\limits_{cyc}\frac{1}{x^3+y^3+xyz}\leq\sum\limits_{cyc}\frac{1}{x^2y+y^2x+xyz}=\sum\limits_{cyc}\frac{z}{x+y+z}=1$,
but I think the first way is better.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Calculating $\int_0^{\pi/3}\cos^2x+\dfrac{1}{\cos^2x}\mathrm{d}\,x$ I've been given the following exercise:
Show that the exact value of $$\int_0^{\pi/3}\cos^2x+\frac{1}{\cos^2x}\,\mathrm{d}x = \frac{\pi}{6}+\frac{9}{8}\sqrt{3}$$
Can someone help me with this?
| This is an answer that requires the knowledge of simple functions such as $\tan$, you can do it without knowing $\sec$ and e.c. Other answers are great, but I am unsure you know all the functions they use.
You can also notice that $\frac{d}{dx}\tan(x)=\frac{1}{\cos(x)^2}$ and that $\cos(x)^2= \frac{1+cos(2x)}{2}$ Thus ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solving $(1+x^2)y' - 2xy = (1+x^2)\arctan(x)$ I'm asked to solve the differential equation: $$(1+x^2)y' - 2xy = (1+x^2)\arctan(x).$$
I rewrite it:
$$y' - \frac{2x}{1+x^2}y = \arctan(x).$$
The integrating factor is:
$$e^{-\int{\frac{2x}{1+x^2}}dx} = e^{-\ln{1+x^2}} = (e^{ln{1+x^2}})^{-1} = \frac{1}{1+x^2}.$$
$$\frac{... | The derivative of $\arctan$ is $x\mapsto \frac{1}{1+x^{2}}$.
Now, observe that for a differentiable function $f$, we have:
$$\int f(x)f'(x)\text{d}x=\frac{1}{2}(f(x))^{2}+K$$
where $K\in\mathbb{R}$. Taking $f=\arctan$, we deduce what you asked.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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What are roots of higher power like $n^9=512$? We know a value of $n$ in $n^2 = 4$ is $\pm2$ (two roots), then what about $n^9 = 512$, generic answer is $+2$, but it should have $9$ roots. What are those $8$ other roots and how can we find them ?
| $$n^9-a^9=(n^3)^3-(a^3)^3=(n^3-a^3)(n^6+a^3n^3+a^6)=(n-a)(n^2+an+a^2)(n^6+a^3n^3+a^6)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Algebra question grade 9 I am a a student and I am having difficulty with answering this question. I keep getting the answer wrong. Please may I have a step by step solution to this question so that I won't have difficulties with answering these type of questions in the future.
Question: Rearrange the formula below to ... | Your first step is correct. So, we have
$$x^2 = \frac{1+n}{1-n}$$
Now, the question is: how do we get rid of this fraction and simplify two $n$'s to one? The natural step is to cross multiply like this:
\begin{align}
{x^2} \cdot \color{#f00}{1} &= \frac{1+n}{1-n} \cdot \color{#f00}1\\
{x^2} \cdot \color{red}{\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
I don't have any idea as to how to proceed with this questio... | From the information provided, we can write
$$ f(x) = p(x)(x-1) + 5,$$
$$ f(x) = q(x)(x+1) + 3,$$
and
$$ f(x) = r(x) (x+2) + 5.$$
where $p(x)$, $q(x)$, and $r(x)$ are polynomials such that
$$\mathrm{gcd} (p(x), x-1 ) = \mathrm{gcd} (q(x), x+1 ) = \mathrm{gcd} (r(x), x+2 ) = 1.$$
Now as
$$ x^3 + 2x^2 -x-2 = (x+1)(x... | {
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"timestamp": "2023-03-29T00:00:00",
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Linear least squares in $\mathbb{R}^{3}$ with three data points. Given data points in the form $(x,y,f(x,y)) = (1,1,7),(1,3,0),(1,-1,8)$, find the least squares solution $\hat x$ to the system of equations $Ax=b$.
Is there enough information to use least squares? The solution created $A$ using the $x$ and $y$ coordinat... | We are given a sequence of measurements $\left\{ x_{k}, y_{k}, f_{k} \right\}_{k=1}^{3}$. Generally, we could hope to find three fit parameters, a trial function like
$$
f(x,y) = c_{0}g_{0}(x,y) + c_{1}g_{1}(x,y) + c_{2}g_{2}(x,y)
$$
where the functions $g$ are linearly independent.
You asked about the linear case w... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Double Integrals - Finding the volume of a unit disk under a function For a Unit Disk
$x^2 + y^2 \ge 1$
And for a function of $x$ and $y$
$f(x, y) = 3 + y - x^2$
I want to find the volume underneath the function bound by the unit disk.
At first I integrated the function with respect to $y$, and made the upper bound $\s... | Note that
$$\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3+y-x^2)dy=\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ydy+\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3-x^2)dy.$$ Now
$$\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ydy=0$$ and $$\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3-x^2)dy=2(3-x^2)\sqrt{1-x^2}.$$
Thus the volume is given by
$$\int_{-1}^1 2(3-x^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061444",
"timestamp": "2023-03-29T00:00:00",
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} |
How can the equation be positive in the interval $ -1 < x < \frac{7-\sqrt{73}}{2}$? I've got the inequality:
$$\frac{x^2-8x-7}{x+1}\geq -1$$
For the inequality to be larger than zero we need either both numerator and denominator to be positive OR both numerator and denominator to be negative.
I draw the number line wit... | It should be as follows :$$\frac { x^{ 2 }-8x-7 }{ x+1 } \geq x+1\\ \frac { x^{ 2 }-8x-7 }{ x+1 } -x-1\ge 0\\ \frac { x^{ 2 }-8x-7-{ x }^{ 2 }-2x-1 }{ x+1 } \ge 0\\ \frac { -10x-8 }{ x+1 } \ge 0\\ \frac { 5x+4 }{ x+1 } \le 0\\ \frac { \left( x+1 \right) \left( 5x+4 \right) }{ { \left( x+1 \right) }^{ 2 } } \le 0\\ \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $3x^2+2\alpha xy+2y^2+2ax-4y+1$ can be resolved into two linear factors, prove that $\alpha$ is the root of the equation $x^2+4ax+2a^2+6=0$.
If $3x^2+2\alpha xy+2y^2+2ax-4y+1$ can be resolved into two linear factors, prove that $\alpha$ is the root of the equation $x^2+4ax+2a^2+6=0$.
I know this question has alrea... | The discriminant of the equation $p(x,y) = 3x^2+(2\alpha y+2a)x +2y^2-4y+1=0$ is $\delta = (2\alpha y+2a)^2-12(2y^2+4y+1)$.
Thus, $$p(x,y) = 3\left(x-\frac{-(2\alpha y+a) +\sqrt{\delta}}{6}\right)\left(x-\frac{-(2\alpha y+a)-\sqrt{\delta}}{6}\right)$$
Now $p(x,y)$ is a perfect square iff the discriminant $\delta_1$ of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that ${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$
If $F_n$ is the $n$-th Fibonacci number ($1,1,2,3,5,8,\dots$), show that
$${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$$
I have tested with a lot of Fibonacci numbers and it seem to obey the ruse, but I don't know ho... | Setting $x=a$, $y=b$, and $z=-(a+b)$ we have to evaluate $\frac{(x^2+y^2+z^2)^2}{x^4+y^4+z^4}$
Since $x+y+z= 0$, $x,y,z$ are roots of $t^3 + t(\sum xy)-xyz = 0$
Hence $x,y,z$ satisfy $t^4 + t^2(\sum xy)-xyzt = 0$
Setting $t=x,y,z$ successively and adding the resulting equations we obtain
$\sum x^4 +\sum x^2 \sum xy -xy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Trying to understand trickier textbook question from less trickier textbook question - number theory Question (difficult):
Show that if $p$ is odd prime, $\binom{p^2}{p} = p \bmod p^3$
Similar question (easier):
Show that if $p$ is odd prime, $\binom{p^2}{p} = p \bmod p^2$
My attempt to answer "easier" question:
$x... | Notice that the calculation which you did actually shows that $$\frac{\binom{p^2}{p}}{p}=\frac{p^2\cdot x}{p^2(p-1)!}=\frac{x}{(p-1)!}\equiv 1\pmod{p^2}.$$ That is, $\frac{\binom{p^2}{p}}{p}-1$ is divisible by $p^2$. Thus $\binom{p^2}{p}-p$ is divisible by $p^3$, so $\binom{p^2}{p}\equiv p\pmod{p^3}$.
In fact, by a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
$\dfrac{2}{\pi} = \dfrac{\sqrt 2}{2} \cdot \dfrac{\sqrt {2+\sqrt 2}}{2} \cdot\dfrac{\sqrt {2+\sqrt {2+\sqrt 2}}}{2} \cdots $ One can show inductively that
$$
\cos \frac{\pi}{2^{n+1}}\ = \frac{\sqrt {2+\sqrt {2+\sqrt {2+\sqrt {\cdots+\sqrt {2 }}}}}}{2},
$$
with $n$ square roots in the right side of the equation.
The se... | Using the trogonometric identity
$$
\sin (2a)=2\sin a\,\cos a\qquad\text{or}\qquad \cos a=\frac{\sin 2a}{2\sin a},
$$
provided that $\,\sin a\ne 0,\,$ we obtain that
$$
\cos(x/2)\cos(x/4)\cdots\cos(x/2^n)=\frac{\sin x}{2\sin(x/2)}\frac{\sin (x/2)}{2\sin(x/4)}\cdots\frac{\sin (x/2^{n-1})}{2\sin(x/2^n)}=\frac{\sin x}{2^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
How to find the projection of $x^2+y^2+z^2+2xyz=1$ on $S^2$? The primary problem is to find the singular points of $g$ when $g$ defines on $S^2$ that $g:S^2 \rightarrow R^3,(x,y,z)\rightarrow(yz-x,zx-y,xy-z)$.
$Note:$The singular point in the problem is the point when the rank of $dg_{p}：T_p S^2 \rightarrow T_{g(p)}R^3... | Let $f \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be defined by the same equation as $g$ so $f(x,y,z) = (yz - x, zx - y, xy - z)$ and $g = f|_{S^2}$. Let $p = (x_0,y_0,z_0) \in S^2$ (so that $x_0^2 + y_0^2 + z_0^2 = 1$). The tangent plane $T_{p}S^2$ can be naturally identified with the plane
$$ T_p S^2 = \{ (x,y,z)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the coefficient of $x^{10}$ in $(1+x^2-x^3)^8$ The coefficient of $x^{10}$ in $(1+x^2-x^3)^8$. I tried to factor it into two binomials but it became way to long to solve by hand.
| $10=2\cdot 5$ or $2\cdot 3+2\cdot 2$. Hence
$$
\binom{8}{5}+\binom{8}{2}\binom{6}{2}=476.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"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.