Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Differentiation of a function We need to find out the derivative $\frac{dy}{dx}$ of the following:
$x^m$$y^n$=$(x+y)^{m+n}$
*
*I know how to differentiate the function and on solving we get $\frac{dy}{dx}$=$\frac{y}{x}$. But we notice that $\frac{dy}{dx}$ is independent of values of $m$ and $n$. So what I did was aga... | Take $log$ on both sides and differentiate w.r.t $x$;
$\frac{m}{x} +\frac{n}{y}. y'=\frac{(m+n)}{(x+y)}. (1+y')$
$\implies y'(\frac{n}{y}-\frac{m+n}{x+y})=\frac{m+n}{x+y}-\frac{m}{x}$
$\implies y'=\frac{y}{x}$ on simplification
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Writing columns of a matrix as linear combinations of other columns Let A be the matrix: $$\begin{pmatrix} 1&2&3&2&1&0\\2&4&5&3&3&1\\1&2&2&1&2&1 \end{pmatrix}$$
What is the best way to write the fifth and sixth columns of the matrix as linear combinations of the first and third columns?
| First consider column 5
We can think of it as
\begin{align}a\begin{pmatrix}1\\2\\1\end{pmatrix}+b\begin{pmatrix}3\\5\\2\end{pmatrix}&=\begin{pmatrix}1\\3\\2\end{pmatrix}\\
&\Downarrow \\
a+3b&=1\\
2a+5b&=3\\
a+2b&=2
\end{align}
So $a=1-3b$
Therefore \begin{align}2(1-3b)+5b&=3\\
2-6b+5b&=3\\
2-3&=b\\
b&=-1
\end{align}
A... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0$? Nine identical balls are numbered $1,2,3,.........,9$ are put in a bag.$A$ draws a ball and gets the number $a$ and puts back in the bag. Next $B$ draws a ball and gets the number $b$.
The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0... | Your solution is correct, but tedious.
A quicker way would be to note that the expression can only be non-positive if $b>5$, which gives only four cases for $b$. In these four cases of $b$, there are $2(b-5)$ values for $a$ that give non-positive results. Therefore the total probability for the expression to be positiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2187290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is $\int_0^{\infty} \frac{x^{p}}{(1+x^2)^3} dx$? We know that
$\int_0^{\infty} \frac{1}{1+x^2} dx = \frac{\pi}{2}.$
(In fact, $F(x)= \int_0^{x} \frac{1}{1+t^2}dt = [\arctan (t)]_0^{x}$, and so $\lim_{x\to \infty} F(x) = \frac{\pi}{2}.$)
Let $p \in \{2,3,4\}, n\in \mathbb N$
How should I evaluate $\int_0^{\infty... | In the integral with $p=3$, the subsitution $y= x^2$ works out well. For $p=2$ and $p=4$ you can use the following partial fractions expansions
$$\frac{x^2}{(1+x^2)^3} = \frac{1}{(1+x^2)^2}-\frac{1}{(1+x^2)^3} $$
$$\frac{x^4}{(1+x^2)^3} = \frac{1}{1+x^2}-\frac{2}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}$$
and the problem is redu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2187684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Showing that $m^2-n^2+1$ is a square Prove that if $m,n$ are odd integers such that $m^2-n^2+1$ divides $n^2-1$ then $m^2-n^2+1$ is a square number.
I know that a solution can be obtained from Vieta jumping, but it seems very different to any Vieta jumping problem I've seen.
To start, I chose $m=2a+1$ and and $n=2b+1$... | I have produced the lemma that rules out negative ratios. It is to be applied after the business of wrting the sum and difference of the pair of odd integers as double new variables.
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
LEMMA
Given integers $$ m > 0, \; \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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$f(x,y) = 0$ when $x = y = 0$. Shouldn't $\frac{\partial^2 f}{\partial y \partial x} = 0$? I'm given the following function:
$f(x,y) =$
$
\begin{cases}
2xy \frac{x^2 - y^2}{x^2 + y^2} & x^2 + y^2 \neq 0 \\
0 & x=y=0
\end{cases}
$
And the task is to prove that $f_{xy}(0,0) = -2$ and that $f_{yx}(0,0) = 2$.
At first glan... | Let $f$ be given by
$$f(x,y)=\begin{cases}
2xy\frac{x^2-y^2}{x^2+y^2}&,x^2+y^2>0\\\\
0&,x^2+y^2=0
\end{cases}$$
For $x^2+y^2>0$, we have
$$\begin{align}\frac{\partial f(x,y)}{\partial x}&=\frac{2y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}\\\\\frac{\partial f(x,y)}{\partial y}&=-\frac{2x(y^4+4x^2y^2-x^4)}{(x^2+y^2)^2}\end{align}$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that there exist integers $p$ and $q$ such that $\det(A^3+B^3) = p^3+q^3$
Let $A$ and $B$ be $3 \times 3$ matrices with integer entries so that $AB = BA$ and $\det(A) = \det(B) = 0$. Prove that there exist integers $p$ and $q$ such that $\det(A^3+B^3) = p^3+q^3$.
I thought about factorizing $A^3+B^3$. We have $... | An alternative, with the additional assumption that both $A$ and $B$ are diagonalizable: Then, since they commute we can write
$$A=PD_AP^{-1}, B=PD_BP^{-1}$$
where $D_A$ and $D_B$ are diagonal matrices. We have
$$A^{3}=PD_A^{3}P^{-1}, B^{3}=PD_B^{3}P^{-1}$$
and
$$det(A^{3}+B^{3})=det(P(D_A^{3}+D_B^{3})P^{-1})=det(D_A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2189377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Three Distinct Points and Their Normal Lines
Suppose That three points on the graph of $y=x^2$ have the property that their normal lines intersect at a common point. Show that the sum of their $x$-coordinates is $0$.
I have a lot going but can not finish it.
Proof:
Let $(a,a^2)$, $(b,b^2)$, and $(c,c^2)$ be three dis... | $b^2a+a^2b=c^2a+a^2c=c^2b+b^2c\rightarrow b^2a+a^2b=c^2a+a^2c$
$\rightarrow b^2a-c^2a=-a^2b+a^2c\rightarrow (b^2-c^2)a=-a^2(b-c)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x^2+\lambda x+1=0, \lambda$ is part of $(-2,2)$ and $4x^3+3x+2c=0$ have common root then $c+\lambda$ can be
If $x^2+\lambda x+1=0, \lambda$ is part of $(-2,2)$ and $4x^3+3x+2c=0$ have common root then $c+\lambda$ can be
I can't figure out a way to solve this question. A hint would be great.
| If $\lambda$ is in $(-2,2)$ (not to be confused with $[-2,2]$), then the roots of $x^2+\lambda x+1=0$ are complex because $x^2+\lambda x +1=0$ only has real solutions when $\lambda=2$ and $\lambda = -2$.
Let $a$ be a shared solution of $x^2+\lambda x+1=0$ and $4x^3+3x+2c=0$. Since $a$ is complex, $4a^3+3a+2c=0$ gives u... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prime number decomposition What is the fastest way to decompose the given number to prime numbers without using calculator?
Example : $$3575$$
What I do is :
$$3575 = 3 \times 10^3 + 5 \times 10^2 + 7 \times 10 + 5 = 3\times5^3\times2^3 + 5^2 \times 2^2 \times 5 + 7 \times 5 \times 2 + 5$$
But now I do not know how to... | By inspection, the final two digits of $75$ means $3575$ is divisible by $25=5^2$. Then the result of dividing out $25$ is $\frac{3500}{25} + \frac{75}{25}=4\times 35+3 = 143=12^2-1 = 11\cdot 13$, giving $3575 = 5^2\cdot11\cdot13$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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When is $\frac{x^2+xy+y^2}{49}$ an integer?
Find the number of distinct ordered pairs $(x, y)$ of positive integers such that $ 1\leq x, y \leq 49$ and $\frac{x^2+xy+y^2}{49}$ is an integer.
Multiplying the given equation by $(x-y)$ gives $x^3 \equiv y^3 \pmod{49}$. Thus we can't have $7 \mid x$ while $7 \nmid y$ or ... | For any given $y$ (not dividing $7$), the congruence $x^3\equiv y^3\pmod{49}$ will have $3$ solutions:
$$x_1=y,\quad x_2=y^{\frac{42+3}3},\quad x_3=y^{\frac{2\cdot42+3}3}\,.$$
(We have $\varphi(49)=42$ possibilities for $y$.)
Since you multiplied by $x-y$, we will have to drop $x_1$ for every $y$, and thus we arrive t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Maximize $P=\frac{\sqrt{a^2-1}}{a}+\frac{\sqrt{b^2-1}}{b}+\frac{\sqrt{c^2-1}}{c} $ Let $a,b,c\geq 1$ satisfy $ 32abc=18(a+b+c)+27$. Find the maximum value $$P=\dfrac{\sqrt{a^2-1}}{a}+\dfrac{\sqrt{b^2-1}}{b}+\dfrac{\sqrt{c^2-1}}{c} $$
$\sqrt {a^2-1}=\sqrt{(a-1)(a+1)}\leq \frac{a-1+a+1}{2}=a$
$\Rightarrow \frac{\sqrt {a... | If $a=b=c=\frac{3}{2}$ so $\sum\limits_{cyc}\frac{\sqrt{a^2-1}}{a}=\sqrt5.$
We'll prove that it's a maximal value.
Indeed, by C-S
$$\sum_{cyc}\frac{\sqrt{a^2-1}}{a}=\sum_{cyc}\sqrt{1-\frac{1}{a^2}}\leq\sqrt{(1+1+1)\sum_{cyc}\left(1-\frac{1}{a^2}\right)}=\sqrt{3\sum_{cyc}\left(1-\frac{1}{a^2}\right)}.$$
Thus, it remains... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sin (\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$, prove that, $\cos [2(\alpha-\beta)]-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$ If $\sin (\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$, prove that, $\cos [2(\alpha-\beta)]-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$.
My Attempt:
.$$\sin (\theta+\alpha)=a$$
$$\sin \theta. \cos \alph... | Consider a circle with radius $\frac{1}{2}$ and a triangle $ABC$ inscribed in the circle with $\angle A = 180^\circ - (\alpha+\theta), \angle B = \theta + \beta$. Then $\angle C = \alpha - \beta$. With the usual notations, we have $a = \sin(\theta+\alpha), b = \sin(\theta+\beta), c= \sin(\alpha - \beta)$. Applying the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$ $$ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$$
My idea for this was to break each numerator into its own fraction as follows
$$ \int_0^1 \left(\frac{1}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{5x^3}{\sqrt{x}}\right)dx$$
$$ \int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2})\... | In derivative we subtract 1.
In integration we add 1.
Your solution is fine. Except last step.
$$=\int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2}) dx $$
$$=\left[2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}\right]_0^1$$
$$=\left[2\cdot1^{1/2} + 2\cdot1^{3/2} + \frac{10}{7}\cdot1^{7/2}\right]-\left[2\cdot0^{1/2} + 2\cdot0^{3/2} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show: $\cos \left( \frac{ 3\pi }{ 8 } \right) = \frac{1}{\sqrt{ 4 + 2 \sqrt{2} }}$ I'm having trouble showing that:
$$\cos\left(\frac{3\pi}{8}\right)=\frac{1}{\sqrt{4+2\sqrt2}}$$
The previous parts of the question required me to find the modulus and argument of $z+i$ where $z=\operatorname{cis{\theta}}$. Hence, I found... | As $\frac{3\pi}{8}$ and $\frac{\pi}{8}$ are complementary angles, we get
$$\begin{align}
\cos\frac{3\pi}{8}&=\sin\frac{\pi}{8}\\
&=\sin\frac{\pi/4}{2}\\
&=\sqrt{\frac{1-\cos(\pi/4)}{2}}\\
&=\sqrt{\frac{1-(1/\sqrt{2})}{2}}\\
&=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}\\
&=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}\cdot \frac{\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Who came up with the identity $a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]$ Though we can prove this it is not something that comes up intutively.
Our ancestors must have been interested in factorising $a^3+b^3+c^3$ but why find it for $a^3+b^3+c^3-3abc$ ?
| Well, it is
$$ \det( aI + b W + c W^2 ) $$
where
$$
W =
\left(
\begin{array}{rrr}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{rrr}
a & b & c \\
c & a & b \\
b & c & a
\end{array}
\right)
$$
Right. It follows that the polynomial is multiplicative: as $W^3 = I$ and $W^4 = W,$ we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
$a_{n+1}=\frac{a_n}{\sqrt{a_n^2+1}}$. Find $a_n$. Given $\{a_n\}$: $a_{n+1}=\frac{a_n}{\sqrt{a_n^2+1}}$, $a_1=1$.
Find $a_n$ like a function of $n$.
My trying.
Let $a_n=\tan\alpha_n$, where $\alpha_n\in\left(0,\frac{\pi}{2}\right)$.
Hence, $a_{n+1}=\sin\alpha_n$ and what is the rest?
Thank you!
| Squaring gives,
$$a_{n+1}^2=\frac{a_n^2}{1+a_n^2}$$
Let $b_n=a_n^2$ then we have,
$$b_{n+1}=\frac{b_{n}}{1+b_{n}}$$
$$=\frac{1+b_n-1}{1+b_n}$$
$$=1-\frac{1}{1+b_n}$$
$$=1-\frac{b_{n+1}}{b_n}$$
$$=\frac{b_{n}-b_{n+1}}{b_{n}}$$
And hence,
$$\frac{b_{n}-b_{n+1}}{b_{n}b_{n+1}}=1$$
Thus,
$$\sum_{n=1}^{N-1} \frac{b_{n}-b_{n+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove this condition if three vectors are colinear? So I was given this problem:
Let $\vec{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\vec{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\vec{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_... | Vectors $\vec{A}$ and $\vec{B}$ are colinear if
$$\vec{A}=k\vec{B}$$
Then the vector product naturally comes out to be $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Finding out the range of $g(x)=\sin x + \cos x$ If $x$ is such that $$\lfloor2\sin x\rfloor+\lfloor{\cos x}\rfloor=-3$$ for some $x \in [0,25]$ then the question is to find out the range of the function $$g(x)=\sin x+\cos x$$Here the function $g(x)$ assumes those values of $x$ which satisfies the above equation in the ... | Note that $\lfloor \cos x \rfloor $ only takes values $-1,0,1$.
Let $S = \{x \in [0,2 \pi] | \lfloor \cos x \rfloor + \lfloor 2\sin x \rfloor = -3 \}$.
If $\lfloor \cos x \rfloor = 0$, then we would need to have $\lfloor 2\sin x \rfloor = -3$ which is impossible.
If $\lfloor \cos x \rfloor = 1$, then we would need to h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Lucas and Fibonacci Numbers Problem:
Let
\begin{align*}
A_0 &= 6 \\
A_1 &= 5 \\
A_n &= A_{n - 1} + A_{n - 2} \; \textrm{for} \; n \geq 2.
\end{align*}
There is a unique ordered pair $(c,d)$ such that $c\phi^n + d\widehat{\phi}^n$ is the closed form for sequence $A_n$.
Find $c$ using the Fibonacci and Lucas number ... | Although the problem is old there are two additional methods that can be considered.
First Method
Given
\begin{align*}
A_0 &= 6 \\
A_1 &= 5 \\
A_n &= A_{n - 1} + A_{n - 2} \; \textrm{for} \; n \geq 2.
\end{align*}
then, since Fibonacci and Lucas numbers satisfy the difference equation, one can take
$$A_{n} = b \, F_{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Evaluating the indefinite integral $\int \frac{x^3}{\sqrt{4x^2 -1}}~dx$.
Find $$\int {x^3\over \sqrt{4x^2 -1}}\,dx.$$
Let
$2x = \sec u$, $2 =\sec (u) \tan(u) u^{'}(x).$ Then
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \, du\\
&= \frac1{16}\int {\sec^4... | Once you get to $\tfrac{1}{16}(\tan u + \tfrac{1}{3} \tan^3 u) + C$, get rid of $u$. Since $\sec u = 2x = 2x/1$, that means $\tan u = \sqrt{4x^2 - 1}$ (draw a right triangle to show this), so $$\begin{aligned}[t]\text{integral} = \tfrac{1}{16}(\tan u + \tfrac{1}{3} \tan^3 u) + C &= \tfrac{1}{16}\Big(\sqrt{4x^2 - 1} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Proving $\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$.
Prove the following identity:
$$\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$$
How can I express $\cos(4\theta) $ in other terms?
| LHS:
$$
\cos (4 \theta)-4\cos(2\theta)=2\cos^2(2\theta)-1-4\cos(2\theta)
$$
RHS:
$$
8\sin^4(\theta)-3=2\left( 2\sin^2 \theta\right)^2-3=2(1-\cos(2\theta))^2-3=2+2\cos^2(2\theta)-4\cos(2\theta)-3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2218085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How do I solve this system of equations (with squares and square roots)? Consider the system of equations:
\begin{align*}
x+y+z&=6\\
x^2+y^2+z^2&=18
\\\sqrt{x}+\sqrt{y}+\sqrt{z}&=4.
\end{align*}
How do I solve this? I've tried squaring, adding equations side by side, substituting, etc., but without success, e.g.
$$x^2+... | \begin{align*}
\text{Let$\,$:}&&a &= \sqrt{x}\\[2pt]
&&b &= \sqrt{y}\\[2pt]
&&c &= \sqrt{z}\\[8pt]
\text{Let$\,$:}&&f(t) &= (t - a)(t - b)(t - c)\\[2pt]
&&&= t^3 - e_1t^2 + e_2t - e_3\\[8pt]
\text{where$\,$:}&&e_1 &= a + b + c\\[2pt]
&&e_2 &= ab + bc + ca\\[2pt]
&&e_3 &= abc\\[8pt]
\text{For $k \in \mathbb{Z}^{+}$, let... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find all possible solutions to $a^2 + b^2 = 2^k$? I'm working on the following problem, this is for an introductory discrete mathematics class.
Find all possible solutions to the equation $a^2 + b^2 = 2^k, k\geq1$ and $a$ and $b$ positive integers.
I've observed that the following are answers:
$2^1 = 1^2 + 1^2$
... | It seems you have at least detected the pattern of the solutions and are on the right path.
A useful fact about squares is that the square of an even number is (of course) a multiple of $4$, whereas the square of an odd number is $1$ plus a multiple of $8$. To see this note that $(2n+1)^2=4\cdot n\cdot(n+1)+1$ and one ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to shorten the derivation of the Laplacian in polar coordinates? Yesterday, I computed the formula for the Laplacian in polar coordinates.
$
\newcommand{p}[2]{\frac {\partial #1} {\partial #2}}
\newcommand{s}[2]{\p {^2 #1} {#2 ^2}}
\newcommand{m}[3]{\p {^2 #1} {#2 \partial #3}}
\newcommand{po}[0]{\p {}}
\newcommand... | Apparently, Math.SE does not allow me to “delete for real” my previous wrong answer, so I will try to provide a non-wrong answer.
Let $\newcommand \t \theta F(x,y) = f(r,\t)$, where $(r,\t) = g(x,y)$. Of course, in our case, $g$ is not a globally well-defined function, but who cares? Differential calculus is all about ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
A Nasty Elliptic Integral I am trying to evaluate:
$$\int_{-\infty}^{\infty}\frac{dx}{\left|1+\alpha x^{2}\right|},\textrm{ }\alpha\in\mathbb{C}\backslash(-\infty,0]$$
It's simple to see that it is convergent for such $\alpha$—but that's probably the only simple thing about it! I've been lost in trying to solve this on... | In this answer I use a variable substitution which I cannot find in the already published answers.
Say that $\alpha \neq 0$ and $\alpha = \varrho e^{i\theta}, \, -\pi <\theta< \pi$. Then $|1+\alpha x^2| = \sqrt{\varrho^2x^4 +2\varrho\cos \theta x^2+1}$ and
\begin{gather*}
I = \int_{-\infty}^{\infty}\dfrac{dx}{|1+\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Probability that $x^2-y^2$ is divisible by $k$ Let two numbers $x$ and $y$ be selected from the set of first $n$ natural numbers with replacement(i.e. the two numbers can be same).The question is to find out the probability that $x^2-y^2$ is divisible by $k\in \mathbb{N}$
For $k=2$
Any number can be expressed as $2p... | A generalization expressed by a set
A good way to generalize this is to use modular arithmetic, or essentially look at the remainder of $\frac{x}{k}$ and $\frac{y}{k}$. As you pointed out in your example for $k=2$, the numbers can only be expressed as $$2*p,2*p+1$$
This can be further generalized into a set of unique e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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How to convert this $k$th term to Telescopic series Given $$t_k=k^4-2k^3+2k^2-1$$ we need to find
$$S=\sum_{k=1}^{k=10}t_k$$
I factorized $t_k$ as
$$t_k=k(k-1)(k^2-k+1)$$ which further i reduced it as
$$t_k=(k^2-k)^2+(k^2-k)$$
Can i have any hint to reduce this in form $$F(k+1)-F(k)$$
| It is more practical to convert $t_k$ into a linear combination of binomial coefficients, then exploit the hockey stick identity. We have
$$ t_k = k^4-2k^3+2k^2-1 = 24\binom{k+1}{4}+6\binom{k}{2}+\binom{k-1}{1} \tag{1}$$
hence:
$$ \sum_{k=1}^{10}t_k = 24\binom{12}{5}+6\binom{11}{3}+\binom{10}{2}=\color{red}{20043}.\tag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I solve $5^{2x} + 4(5^x) - 5 = 0$? This is a math problem I'm currently working on.
$$5^{2x} + 4(5^x) - 5 = 0$$
I've used logarithm to try solve the problem. Here's what I've done so far:
\begin{align}5^{2x} + 4(5^x) - 5 &= 0\\
5^{2x} + 5^x &= \frac{5}{4}\\
\log_5{2x} + \log_5{x} &= \log_5\left(\frac{5}{4}\rig... | You made a pretty big mistake because $\log a + \log b \neq \log (a+b)$. So, your third line does not follow from your second.
To actually solve the problem, here's two hints:
*
*Set $y=5^x$
*Use the fact that $a^{bc} = (a^b)^c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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what is the relationship between the SVD of $[A;B]$ and that of $A$ and $B$? In my question, $A\in R^{m\times r}$, $B\in R^{n\times r}$, and $[A;B]\in R^{(m+n)\times r}$ results from stacking $A$ over $B$. Given $m>r$ and $n>r$, we do the SVD on $A$ and $B$ and have that $A=U_aS_aV_a^T$ and $B=U_bS_bV_b^T$, where $U_a\... | Not really, even take the simplest case of $1 \times 1$ matrices.
*
*Let $A= \begin{bmatrix}1 \end{bmatrix}$ and let $B=\begin{bmatrix} 2 \end{bmatrix}$.
*Singular value of $A$: $\sigma_a = 1$
*Singular value of $B$: $\sigma_b = 2$
*SVD of $[A ; B] = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$: $$U S V^T = \begin{bma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2228930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove inequality $\sum\frac{{{a^3} - {b^3}}}{{{{\left( {a - b} \right)}^3}}}\ge \frac{9}{4}$ Given $a,b,c$ are positive number. Prove that $$\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge \frac{9}{4}$$
$$\Leftrightarrow \sum \frac{3(a+b)^2+(a-b)^2}{(a-b)^2} ... | your inequality is true, since it is equivalent to
$$3/4\,{\frac { \left( {a}^{2}b+{a}^{2}c+a{b}^{2}-6\,abc+a{c}^{2}+{b}^{2
}c+b{c}^{2} \right) ^{2}}{ \left( -c+a \right) ^{2} \left( b-c
\right) ^{2} \left( a-b \right) ^{2}}}
\geq 0$$ which is obviously true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
solve for $x$ and $y$ in the following equation $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$ Solve for $x$ and $y$ in the following equations: $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$. I made $y^2$ the subject of the formula in eqn 2.
This gives $y^2 = 6 -2x^2$.
I substitute this into the first eqn.
This gives $x^2+3xy -3$. There'... | If
$$
\begin{cases}
2x^2+y^2=6\\
x^2+y^2=3+3xy \\
\end{cases}
$$
then by setting $x^2=a$, $y^2=b$
$$
\begin{cases}
2a+b=6 \\
a+b=3+3xy \\
\end{cases}
$$
Hence, by Cramer's rule (of linear algebra)
$$
\begin{cases}
a=x^2=3-3xy & (1)\\
b=y^2=6xy & (2)
\end{cases}
$$
Hence, by (2) $y=0$ or $y=6x$. P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Change of coordinates in a polynomial ring I want to prove that $$zy^2+z^2y+bxyz+x^3+cz^3+dx^2z+az^2x\in \mathbb{C}[x,y,z]$$ can be transformed into $$y^2z+x^3+\alpha xz^2+\beta z^3$$ by means of an invertible linear change of coordinates.
I tried to complete the square in $y$, I got $z(y^2+yz+xy)=z(y+t)^2-zt^2$ where ... | Rewrite $$zy^2+z^2y+bxyz+x^3+cz^3+dx^2z+az^2x=0$$ as $$y^2z+bxyz+yz^2=x^3+dx^2z+axz^2+cz^3.$$ Make the change of coordinates $y$ goes to $y-\frac{bx+z}{2}$. This gives $$y^2z-\frac{bxz^2}{2}-\frac{b^2x^2z}{4}-\frac{z^3}{4} = x^3+dx^2z+axz^2+cz^3$$
Or equivalently, after collecting terms, $$y^2z=x^3+\left(d+\frac{b^2}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the variance of sum of indicator variables
Compute $\operatorname{Var} (X)$ where $X = \sum_{i=1}^n x_i$
I am aware of the formula
$$\operatorname{Var} (X) = \sum_{i=1}^9 \operatorname{Var} (X_i) + \sum_{i \ne j} \operatorname{Cov } (X_i, X_j)$$
But I cannot seem to apply it here
Edit:
We know that $\text{Var }... | \begin{align}
X_i & = \begin{cases} 1 & \text{if the $i$th and $(i+1)$th characters are LW or WL}, \\ 0 & \text{if LL or WW}, \end{cases} \\[10pt]
& = \begin{cases} 1 & \text{with probability } \left(\dfrac 1 4 \times \dfrac 3 4\right) + \left(\dfrac 3 4\times \dfrac 1 4 \right) = \dfrac 3 8, \\[8pt] 0 & \text{with pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Why $\frac {a}{b} = \frac {c}{d} \implies \frac {a+c}{b+d} = \frac {a}{b} = \frac {c}{d}$? In geometry class, this property was quickly introduced:
$$\dfrac {a}{b} = \dfrac {c}{d} \implies \dfrac {a+c}{b+d} = \dfrac {a}{b} = \dfrac {c}{d} $$
It usually avoids some boring quadratic equations, so it's useful but I don'... | $$\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a+c}{b+d}=\frac{a+\frac{ad}{b}}{b+d}=\frac{ab+ad}{b(b+d)}=\frac{a}{b}=\frac{c}{d}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Limit of $\int _{\frac{1}{n}}^n \frac{\arctan x}{x^2 + 2ax + 1}\,\mathrm{d}x$ without Taylor expansion $\def\d{\mathrm{d}}$Consider $$I_n(a) = \int _{\frac{1}{n}}^n \frac{\arctan x}{x^2 + 2ax + 1}\,\d x, \quad n\ge 1,\ a \in [0, 1)$$
and evaluate
$$\lim _{n \to \infty} I_n(a). \quad a \in (0, 1)$$
One thing I have foun... | You are almost there. Since
$$
I_n(a)=\int_{1/n}^n\frac{\arctan x}{x^2+2ax+1}\,dx
=\int_{1/n}^n\frac{\arctan (1/x)}{x^2+2ax+1}\,dx
$$
and
$$
\arctan x+\arctan(1/x)=\frac\pi2
$$
you find that
$$
2I_n=\int_{1/n}^n\frac{\arctan x+\arctan(1/x)}{x^2+2ax+1}\,dx
=\int_{1/n}^n\frac{\pi/2}{x^2+2ax+1}\,dx.
$$
Hence
$$
I_n(a)=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How to solve the differential equation $y' = \frac{x+y}{x-y}$? Solve the following differential equation:
$y' = \frac{x+y}{x-y}$
Someone please help me start this problem. This does not look like a regular first-order differential equation in the form $y' + 2xy = x$. Thank you.
| Hint: It's homogenous of degree 1 which suggests the substitution
$$
y=xv\implies y'=xv'+v
$$
and
$$
xv'+v=\frac{x+xv}{x-xv}=\frac{1+v}{1-v}\stackrel{\text{algebra}}{\implies}
\frac{1-v}{v^2+1}\mathrm dv=\frac{1}{x}\mathrm dx\implies \int \frac{1-v}{v^2+1}\,\mathrm dv=\ln x+c\\
\implies \int\frac{1}{v^2+1}\,\mathrm d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Let $( \sqrt{2} + 1)^{1000} = a + b \sqrt{2}$, where a and b are integers. Find the greatest common factor of b and 81. Let $( \sqrt{2} + 1)^{1000} = a + b \sqrt{2}$, where $a$ and $b$ are integers. Find the greatest common factor of $b$ and $81$.
How would one solve this question? I tried to use the binomial theorem b... | This is so long for a comment:
We say $ ( \sqrt{2} + 1)^{n} = a_n + b_n \sqrt{2}$ and $a_n,b_n$ are integers. Therefore $( 1- \sqrt{2})^{n} = a_n - b_n \sqrt{2}$. For positive odd value of $n=2k-1$, by product
$-1=( \sqrt{2} + 1)^{n} ( 1- \sqrt{2} )^{n}= (a_n + b_n \sqrt{2})(a_n - b_n \sqrt{2})$.
Hence $a_n^2 - 2b_n^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Least possible polynomial deegree of complex roots What is the least possible deegree of polynomial with real coefficients having roots $2\omega , 2+3\omega , 2+3\omega ^2 , 2-\omega -\omega ^2$
As there are four roots so the deegree should be four but the answer is given as five . how ?
| Let's look at the given roots. We see that one of these roots,
$2-\omega -\omega ^2$, is in fact real:
$$
2-\omega -\omega ^2 = 3, \quad\mbox{ while }\quad
(x-(2+3\omega))(x-(2+3\omega^2)) = x^2-x+7.
$$
A fourth degree polynomial with the four given roots would have complex coefficients, namely, it's the polynomial
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $e^x>x(x+1)$
Let $x>0$,prove that $$e^x>x(x+1)$$or
$$x>\ln{x}+\ln{(x+1)}$$
we can use this Taylor some first four term,
$$e^x>1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3$$ prove it
But this inequality it seem very nice,maybe there exsit simple methods or Amazing way?
| We need to prove$$e^x>1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3\ge x^2+x$$which is$$x^3+3x^2+6x+6\ge 6x^2+6x$$which is$$x^3-3x^2+6=(x-2)^2(x+1)+2>0$$which is obvious for positive $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How do you find the Taylor Series of $f(x) = x^{2}*\sin(\frac{x^{2}}{3})$ about $a = 0$? Please without getting too technical, this is calculus 2. I know that the general form of a Taylor Series about 0 (Maclaurin) is: $$\sum_{n=0}^{\infty} \frac{f^{n}(0)}{n!} * x^{n}$$ and so, $$f(x) = f(0)/0! + f (0)/1!*x + f (0)/2!*... | Use Taylor Series for $\sin x$
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!} -\frac{x^7}{7!}\ldots$$
$$\sin {\left(\frac{x^2}{3} \right)}=x-\frac{{\left(\frac{x^2}{3} \right)}^3}{3!}+\frac{{\left(\frac{x^2}{3} \right)}^5}{5!} -\frac{{\left(\frac{x^2}{3} \right)}^7}{7!}\ldots$$
You want $x^2 \sin {\left(\frac{x^2}{3} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove by induction that $\forall n\geq 1,\ 7\mid 3^{2n+1} + 2^{n-1}$ Prove by induction that $$7 \mid 3^{2n+1} + 2^{n-1},\ \forall n\geq 1$$
Base case $n=1$:
$$3^{2 × 1+1} + 2^{1-1} = 28.$$
Induction:
$$P(k): 3^{2k+1} + 2^{k-1},\ P(k+1): 3^{2(k+1)+1} + 2^{(k+1)-1}.$$
$$3^{2k+3} + 2^k = 9 \times 3^{2k+1} + 2^{k-1} \tim... | By the binomial theorem
$$
3^{2n+1}=3\cdot 9^n=3(7+2)^n=3(7a+2^n)=21a+6\cdot2^{n-1}
$$
Therefore
$$
3^{2n+1}+2^{n-1}=21a+7\cdot2^{n-1}=7(3a+2^{n-1})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Radii of inscribed and circumscribed circles in right-angled triangle In a right angled triangle, △ ABC, with sides a and b adjacent to the right angle, the radius of the inscribed circle is equal to r and the radius of the circumscribed circle is equal to R.
Prove that in △ABC, $a+b=2\cdot \left(r+R\right)$.
| We have the following two formulas for $R$ and $r$:
*
*$$R=\tfrac12c=\tfrac12 \sqrt{a^2+b^2}$$
($c^2=a^2+b^2$ by Pythagoras.)
*
*$$r=\dfrac{ab}{a+b+c}$$
(proof: the area of the (right) triangle ABC, i.e., $\tfrac12ab$ is equal to the sum of the areas of triangles $IBC, ICA, IAB$, i.e., $\tfrac12ra+\tfrac12r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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If $\frac{a}{b} = \frac{c}{d}$ why does $\frac{a+c}{b + d} = \frac{a}{b} = \frac{c}{d}$? Can anyone prove why adding the numerator and denominator of the same ratios result in the same ratio? For example, since $\dfrac{1}{2}=\dfrac{2}{4}$ then $\dfrac{1+2}{2+4}=0.5$.
| Consider $\frac{a}{b}=\frac{ka}{kb}$ Then, $$\frac{a+ka}{b+kb}=\frac{(k+1)a}{(k+1)b}=\frac{a}{b}$$
which is exactly what you noticed, but with $a=1,b=2,k=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 1
} |
bijection between number of non-congruent triangles with perimeter 2n and integer sides and number of partitions of n into exactly three terms. Is there any a bijection between the following problems?
Let $f(n)$ be the number of non-congruent triangles with perimeter 2n
and integer sides. Let $h(n)$ be the number of pa... | The number of non congruent triangles is given here ... https://oeis.org/A005044
The first few values are $0, 0, 0, \color{red}{1}, 0, \color{red}{1}, 1, \color{red}{2}, 1, \color{red}{3}, 2, \color{red}{4}, 3, \color{red}{5}, 4,\color{red}{7}, 5, \color{red}{8}, 7, \color{red}{10}, 8, \color{red}{12}, 10, \color{red}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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how SVD is calculated in reality let us suppose that we have following matrix
$
A=
\left[ {\begin{array}{cc}
4 & 0 \\
3 & -5 \\
\end{array} } \right]
$
for calculation of SVD,first i have calculated $A'*A$ which is equal to
$
A'*A=
\left[ {\begin{array}{cc}
25... | You did everything correctly. You already found $\sigma_1,\sigma_2$ - they are the square roots of the eigenvalues of $A^TA$ so $\sigma_1 = \sqrt{40},\sigma_2 = \sqrt{10}$. After you found $V$ with columns $v_1,v_2$, you must have
$$ Av_1 = \sigma_1 u_1, Av_2 = \sigma_2 u_2 $$
so $u_1$ is just $\frac{Av_1}{\sigma_1}$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Polynomials $P(x)$ such that $P(x-1)$ $=$ $P(-x)$ Are there an infinite number of polynomials $P(x)$ of even degree such that
$P(x-1) = P(-x)$
$P(x) = x^2+x+1$ is a good example because $P(x-1) = (x-1)^2+(x-1)+1 = x^2-x+1 = P(-x)$.
$P(x) = x^4+2x^3+4x^2+3x+1$ is another example because $P(x-1) = (x-1)^4+2(x-1)^3+4(x-1... | The condition $P(-x)=P(x-1)$ is equivalent to $P(x-1/2)=P(-1/2-x)$,
namely that $P(x-1/2)$ is an even function. That should be helpful.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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For how many numbers $X^2 \equiv X \mod 10^n$? I've been looking through this post:
Square of four digit number $a$
And wanted to see if there's a way to generalize the idea, in order to answer the following question:
For how many numbers $X^2 \equiv X \mod 10^n$?
Also, is $9376$ the only four digit number such tha... | The number of distinct numbers modulo $10^n$ solving
$$x^2\equiv x\pmod{10^n}\tag1$$
is four.
To see this by the Chinese remainder theorem $x$ solves (1) iff both
$$x^2\equiv x\pmod{2^n}\tag2$$
and
$$x^2\equiv x\pmod{5^n}\tag3.$$
These congruences state that $p^n\mid x(x-1)$ for $p=2$ or $5$.
As $x$ and $x-1$ have no c... | {
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Find a quadratic integer polynomial that annihilates a given $2\times2$ integer matrix Suppose $A=\begin{bmatrix}3&2\\2&3\end{bmatrix}$. How can we find integers $b$ and $c$ such that $A^2+bA+cI_2=0$?
| This is an application of the Cayley-Hamilton Theorem.
Every square matrix satisfies its own characteristic function.
Compute the characteristic polynomial of
$$\mathbf{A}=
\left(
\begin{array}{cc}
3 & 2 \\
2 & 3 \\
\end{array}
\right)
$$
$$ \det \mathbf{A} = 5, \qquad \text{tr }\mathbf{A} = 6$$
The characteristic... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding rank of a matrix depending on parameter a We have the matrix
$$A = \begin{bmatrix}1 & a & 1\\-4 & 2 & a \\ a & -1 & -2 \\ 4 & 6 & 1\end{bmatrix}$$
and we want to find the rank of this matrix as a function of parameter $a$.
| The rank of the matrix is equal to the number of nonzero rows in the matrix after reducing it to the row echelon form using elementary transformations over the rows of the matrix.
$\ \begin{bmatrix}1 & a & 1\\-4 & 2 & a \\ a & -1 & -2 \\ 4 & 6 & 1\end{bmatrix}$ substract (-4)*row 1 from row 2 ->
$\ \begin{bmatrix}1 & a... | {
"language": "en",
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If $4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5 $ when divided by $(x+1)$ gives a remainder of -14, then the value of k equals? If $4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5 $ when divided by $(x+1)$ gives a remainder of -14, then the value of k equals?
I got this and similar type of question in a book and I don'... | Let $f(x)=4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5 $. Then $f(-1)=-14$ since the remainder when we divide by $x+1$ is $-14$.
So, $4+1-3-5+k-2+1+k-5-5=-14$
Thus, $2k=4$and $k=2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show that $(n+1)(n+2)...(2n)$ is divisible by $2^n$ but not by $2^{n+1}$ Is this proof correct?
Suppose $2^k$ is the largest power of $2$ in the sequence $n+1, n+2, ... 2n$
Then we can compute the power of 2 in the product as $n/2 + n/2^2 + ... n/2^k = n(1 + 2 + .... 2^{k-1})/2^k = n$.
| Note that
$$\dfrac{(n+1)(n+2)\dots (2n)}{1\cdot 3 \cdot 5 \dots (2n-1)} = \dfrac{(2n)!/n!}{(2n)!/(2\cdot 4 \cdot 6 \cdot \dots \cdot (2n))} = \dfrac{(2n)!/n!}{(2n)!/(2^n n!)} = 2^n$$
Hence,
$$(n+1)(n+2)\dots (2n) = 1\cdot 3 \cdot 5 \dots (2n-1) \cdot 2^n$$
Since the first $n$ multipliers on the right side are all odd, ... | {
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How to find two vectors orthogonal to the gradient space of the feasible set? After finding the solution to the minimization of $f(x,y,z)=\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\left(z-\frac{1}{2}\right)^2$ constrained by $h(x,y,z)=x^2+y^2+z^2-1=0$ by using the Lagrange multiplier method:
$$\nabla f(... | I guess that the idea is that you can choose any two vectors, as long as they are independent to $\nabla h$, but this is easy to achieve by simple inspection, e.g. two vectors from the canonical basis $e_1=(1,0,0)$, $e_2=(0,1,0)$ will do the job. Then apply Gram-Schmidt to the basis $(\nabla h,e_1,e_2)$ and you obtain ... | {
"language": "en",
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Integrate $\int \frac {1}{(x+2)(x+3)} \textrm {dx}$ Integrate $\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$
My Attempt:
$$\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$$
$$\int \dfrac {1}{x+2} \textrm {dx} . \int \dfrac {1}{x+3} \textrm {dx}$$
$$\dfrac {\textrm {log (x+2)}}{1} . \dfrac {\textrm {log (x+3)}}{1} + C$$
$$\textrm ... | Method to do -
$\frac{1}{(x+3)(x+2)} =\frac{A}{x+2}+\frac{B}{x+3}$
$1 = A(x+3)+B(x+2)$
Case 1 -
When $x+3=0$
$x=-3$
Put $x=-3$
$1 = -B$
$B = -1$
Case 2 -
When $x+2=0$
$x=-2$
Put $x=-2$
$1 = A$
$A = 1$
Now your integral becomes,
$\int (\frac{1}{x+2}-\frac{1}{x+3})\,dx$
| {
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Finding all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ Find all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ for all $x>0$
Attempt: Let $$\sqrt{9-(x-a)^2}> \sqrt{16-x^2}$$
So $$x^2-(x-a)^2>7$$
So $$a(2x-a)>7\Rightarrow a^2-2ax+7<0$$
could some help me how to find range of $a,... |
The functions
$g(x;a) = \sqrt{9-(x-a)^2}$ and
$f(x)=\sqrt{16-x^2}$
are semicircles whose domains are $x \in [a-3, a+3]$ and $x \in [-4,4]$ respectively. Pictured above are $f(x), g(x;-1)$ and $g(x;1)$.
As the above picture suggests, $\sqrt{9-a^2+2ax-x^2} \le \sqrt{16-x^2}$ for all $x \in [-1,1]$ when $a \in [-1,1... | {
"language": "en",
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How does $y=\frac{2x+2}{x+2}$ contribute to demonstrating the Dedekind cut for $\sqrt{2}$? In a previous question a respondent corrected my description a Dedekind cut of $\sqrt{2}$. This Wikipedia article states: "Showing that it is a cut requires showing that for any positive rational $x$ with $x^2 < 2$, there is a ra... | Let $x$ be a positive rational number with $x^2<2$. First note that
$$x < \frac{2x+2}{x+2}$$
is equivalent to
$$x^2+2x<2x+2$$
which in turn is equivalent to $x^2<2$, which is true by hypothesis. So $x<y$.
Now note that
$$\left( \frac{2x+2}{x+2} \right)^2< 2$$
is equivalent to
$$(2x+2)^2 < 2(x+2)^2$$
which simplifies ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding minimum of trigonometric function Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got max... | By C-S we obtain:
$$v=$$
$$=\sqrt{a^2\cos^2x+b^2\sin^2x+b^2\cos^2x+a^2\sin^2x+2\sqrt{(a^2\cos^2x+b^2\sin^2x)(b^2\cos^2x+a^2\sin^2x)}}=$$
$$=\sqrt{a^2+b^2+2\sqrt{(a^2\cos^2x+b^2\sin^2x)(b^2\cos^2x+a^2\sin^2x)}}\geq$$
$$\geq\sqrt{a^2+b^2+2|ab|(\cos^2x+\sin^2x)}=|a|+|b|.$$
$$v\leq\sqrt{2(a^2\cos^2x+b^2\sin^2x+b^2\cos^2x+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $x^{2^{n+1}} - x^{2^n} + 1$ is irreducible over the integers for all $n$? It seems $x^2-x+1$ or $x^4-x^2+1$ are irreducible over the integers.
Is $x^{2^{n+1}} - x^{2^n} + 1$ is irreducible for all non-negative integer $n$? If so, how to prove?
| Multiplying by $ (X^{2^n} + 1) $, we have that
$$ (X^{2^n} + 1)(X^{2^{n+1}} - X^{2^n} + 1) = X^{3 \cdot 2^n} + 1 $$
It is clear from this that the roots of $ X^{2^{n+1}} - X^{2^n} + 1 $ are the primitive $ 3 \cdot 2^{n+1} $th roots of unity, and the degree of the corresponding minimal polynomial is $ \varphi(3 \cdot 2... | {
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Integrals with fractions in exponentials I came across this integral in a research paper while trying to understand Bernstein's inequalities. I want to upper bound or evaluate this integral but it looks too complicated. The integral is:
\begin{align}
I=\int_0^\infty \exp \left ( \frac{-at^2}{bt+c} \right ) dt, \quad a,... | $\int_0^\infty e^\frac{-at^2}{bt+c}~dt$
$=\int_c^\infty e^\frac{-a\left(\frac{t-c}{b}\right)^2}{t}~d\left(\dfrac{t-c}{b}\right)$
$=\dfrac{e^\frac{2ac}{b^2}}{b}\int_c^\infty e^{-\frac{at}{b^2}-\frac{ac^2}{b^2t}}~dt$
$=\dfrac{e^\frac{2ac}{b^2}}{b}\int_1^\infty e^{-\frac{act}{b^2}-\frac{ac^2}{b^2ct}}~d(ct)$
$=\dfrac{ce^\f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is there a possible explanation in plain English how to use the Chinese Reminder Theorem? For example, if it is the problem of
Find the smallest integer that leave a remainder of 3 when divided by
5, a remainder of 5 when divided by 7, and a remainder of 7 when
divided by 11
In some explanation such as in this ar... |
Find the smallest integer that leave a remainder of 3 when divided by 5, a remainder of 5 when divided by 7, and a remainder of 7 when divided by 11
We want to find a solution to:
$$
\begin{align}
x&\equiv 3 \pmod 5\\
x&\equiv 5 \pmod 7\\
x&\equiv 7 \pmod {11}
\end{align}
$$
The solution is unique $\pmod {385}$.
We c... | {
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Find the value of $\sum_{n=1}^{\infty} \frac{2}{n}-\frac{4}{2n+1}$ Find the value of $$S=\sum_{n=1}^{\infty}\left(\frac{2}{n}-\frac{4}{2n+1}\right)$$
My Try:we have
$$S=2\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{2}{2n+1}\right)$$
$$S=2\left(1-\frac{2}{3}+\frac{1}{2}-\frac{2}{5}+\frac{1}{3}-\frac{2}{7}+\cdots\right... | We can use the digamma function $\psi$. It is known that
$$\psi(z+1)=-\gamma+\sum_{n=1}^{\infty}\dfrac{z}{n(n+z)},\ \ \forall z \in \mathbb{C} \setminus \{-1,-2,...\}.$$
Thus for $z=\frac{1}{2}$ we obtain
$$\psi\left(\dfrac{3}{2}\right)=-\gamma+\sum_{n=1}^{\infty}\dfrac{1}{n(2n+1)}=-\gamma+\sum_{n=1}^{\infty}\left(\dfr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim _ { x \rightarrow 0} \frac { \tan ( x ^ { a } ) - ( \sin x ) ^ { a } } { x ^ { a + 2} }$ For $a \geq 1$, evaluate $\lim _ { x \rightarrow 0} \frac { \tan ( x ^ { a } ) - ( \sin x ) ^ { a } } { x ^ { a + 2} }$
I would like to solve this without Taylor's expansion. I've tried adding and subtracting $x^a$ b... | If $a$ is a positive integer then there is an easy approach via factorization. Let us first deal with $a=1$. We have $$\frac{\tan x - \sin x}{x^{3}} =\frac{\sin x} {x} \frac{1}{\cos x} \frac{1-\cos x} {x^{2}}\to \frac{1}{2}\tag{1}$$ Also note that using L'Hospital's Rule or Taylor series we can easily show that $$\frac... | {
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Convolution Integral problem I'm having a hard time to do this exercise. I'm using the definition for convolution but I'm stuck at the integral. Thanks in advance.
For $t>0$ consider $f_a(x)=\frac{1}{\sqrt{4 \pi a}}e^{-\frac{x^2}{4a}}$.
Show that $f_a*f_b=f_{a+b}$
By definition: $(f*g)(x)$= $\int_{-\infty}^{+\infty} ... | One of the most useful elementary tricks to know when dealing with Gaussian-type integrals is completing the square.
On that note, we calculate
\begin{align}
f_a*f_b(x)
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{y^2}{4a}-\frac{(x-y)^2}{4b}\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve the first order differential equation, $\frac{x\mathop{dx}-y\mathop{dy}}{x\mathop{dy}-y\mathop{dx}}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$ I need to find the general solution to the differential equation,
$$\dfrac{x\mathop{dx}-y\mathop{dy}}{x\mathop{dy}-y\mathop{dx}}=\sqrt{\dfrac{1+x^2-y^2}{x^2-y^2}}$$
I have tried si... | Substitute $x=r \sec \theta$, $y=r \tan \theta$.
$x^2 - y^2 = r^2$
Taking differentials, $xdx - y dy = r dr$
and $\frac y x=\sin\theta$
Taking differentials, $\dfrac {x dy - y dx} {x^2} = \cos \theta d\theta$
or, $x dy - y dx = r^2 \sec \theta d\theta$
Thus the differential equation becomes
$\dfrac {rdr}{r^2 \sec\the... | {
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Show that $x^{12}-x^9+x^4-x+1\geq0$ for all $x$ Show that $x^{12}-x^9+x^4-x+1\geq0$ for all $x$.
When $x\leq0$ , this is easy.
When $x\geq1$, then also this is easy.
I need help with the case when $0\leq x\leq 1$
| $x^{12} - x^9 +x^4 - x + 1 \ge 0$;
Rewritten:
$x^{12} + (x^4 - x^9) + (1 - x) \ge 0$.
For $0 \le x \le 1$:
We have $x^{12} \ge 0$; $(x^4 -x^9) \ge 0$ and $(1-x) \ge 0$.
The sum of the above terms is $ \ge 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Divisors of the form $4n+1$ I read a question
Number of divisors of the form $(4n+1)$
.
In the soultion
Any positive divisor of $2^2\cdot 3^3\cdot 5^3\cdot 7^5$ of the form
$4k+1$ is a number of the form: $$3^a\cdot 5^b\cdot 7^c$$ with $0\leq
> a\leq 3,0\leq b\leq 3,0\leq c\leq 5$ and $\color{red}{a+c}$ being even. ... | HINT:
As $3=4-1,7=8-1$
$$3^a7^c=(4-1)^a(8-1)^b$$
using Binomial Expansion, we can prove it to be of the form $4c+(-1)^{a+b}$ where $c$ is some integer
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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To prove $0.4 \le \int_{0}^{1} x^{\sin x+\cos x} dx \le 0.5$ Prove that $$0.4 \le \int_{0}^{1} x^{\sin x+\cos x} dx \le 0.5$$
i have proved as follows:
when $x \in (0 \: \: 1)$ $$\sin x+\cos x \in (1 \:\:\:\sin 1+\cos 1)$$
and so $$x^{\sin x+\cos x} \in (x^{\sin 1+\cos 1} \:\:\: x)$$ $\implies$
$$\int_{0}^{1}x^{\sin x... | You may note that $\sin x+\cos x =\sqrt 2\sin(x+\frac \pi 4)$.
By investigating the monotonicity of $\sin x$, we have, $\forall x \in [0,1]$, $\sin x+\cos x =\sqrt 2\sin(x+\frac \pi 4) \in [1, \sqrt 2]$.
Now can you go further starting from here?
P.S: To get the first identity, we can exploit the sine of the sum ident... | {
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For $x \ge 1$, is $2^{\sqrt{x}} \ge x$ For $x \ge 1$, is $2^{\sqrt{x}} \ge x$
The answer appears to me to be yes since:
$$\sqrt{x} \ge \log_2 x$$
Here's my reasoning:
(1) For $x > 16$, $5\sqrt{x} > 4\sqrt{x} + 4$
(2) For $x > 25$, $x > 5\sqrt{x}$
(3) So, $x > 25$, $2x > x + 4\sqrt{x} + 4$ and it follows that $\sqrt{2x}... | The claim is in fact false. Take $x=9$, we have
$$2^{\sqrt{9}}=2^3=8<9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$. If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
$$\sec \theta + \tan \theta = x$$
$$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$
$$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$
$$1+\sin \theta =... | WLOG let $\theta=\dfrac\pi2-2y\implies x=\csc2y+\cot2y=\dfrac{1+\cos2y}{\sin2y}=\cot y$
$$\sin\theta=\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=\dfrac{\cot^2y-1}{\cot^2y+1}=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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A number $x$ such that $x^n = x$ for every $n$ Is there proof fully algebraic that shows that $1$ is the only number that when put in the function $f(x) = x^n$, for $n = $ any number, the output is always $x$?
This is pretty obvious, but you seem to prove it using the trivial knowledge .
For example, $1^4 = 1, 1^{463... | For positive integer $n$, write
$$f(x)=x^n=x^n-1+1=(1+x+\cdots+x^{n-1})(x-1)+1.$$
If $x > 1$, then $(1+x+\cdots +x^{n-1})(x-1) > n(x-1) $ so $f(x) > n(x+1)+1 > 1$.
If $0<x < 1$, then $(1+x+\cdots +x^{n-1})(x-1) < \frac{1}{1-x}(x-1) = -1$. Hence $f(x) < -1+1 = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
What is the limit of this sequence? How does it relate to the exponential function?
Find $$\lim_{n\to\infty} \left(1+\frac{x^2}{n^2}\right)^n$$
I particular, I am hoping to find the above to be $1+x^kg(x/n)$ where $g(\cdot)$ is a function with uniformly bounded derivatives.
Edit:
I'm trying to prove a result that wou... | The limit can be evaluated without appeal to the exponential function. To proceed we first note that
$$\left(1+\frac{x^2}{n^2}\right)^n\left(1-\frac{x^2}{n^2}\right)^n\le 1\tag1$$
Then, rearranging $(1)$, we obtain
$$ \left(1+\frac{x^2}{n^2}\right)^n\le \frac{1}{\left(1-\frac{x^2}{n^2}\right)^n}\tag 2$$
Using Bernoul... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a,b,c,d$ are real numbers such that....
Suppost that $a,b,c,d$ are real numbers such that $$a^2+b^2=1$$ $$c^2+d^2=1$$ $$ac+bd=0$$
I've to show that $$a^2+c^2=1$$ $$b^2+d^2=1$$ $$ab+cd=0$$
Basically,I've no any idea or tactics to tackle this problem. Any methods? Thanks in advance.
EDITED.
The given hint in the book... | Just a very basic brute force method,
$a^2 + b^2 = 1$ - (1)
$c^2 + d^2 = 1$ -(2)
$ac+bd=0$
On multiplying the two we get, $a^2c^2+b^2d^2+a^2d^2+b^2c^2=1$
Whereas $a^2c^2+b^2d^2+2abcd=0$ {squaring both sides $ac + bd = 0$}
Thus we get $-2abcd+a^2d^2+b^2c^2=1 \Rightarrow (ad-bc)^2=1 \Righ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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$\frac{'ab'}{'ba'}=\frac{'bc'}{'cb'}$ Question:
We write $'pq'$ to denote a two digit integer with tens digit $p$ and units digit $q$. For which values of $a,b$ aand $c$ are the two fractions $\frac{'ab'}{'ba'}$and$\frac{'bc'}{'cb'}$ equal and different from $1$?
My attempt:
So, write this using "normal algebra";$... | Looks good so far. Now you just have to write down all the ways that $b^2 = ac$ for $1 \leq a, b, c \leq 9$, $a \neq b$, $b \neq c$ : $$\begin{align*} 2^2 &= 1 \times 4 \\ 3^2 &= 1 \times 9 \\ 4^2 &= 2 \times 8 \\ 6^2 &= 4 \times 9 \end{align*} $$
Each of these gives two solutions; e.g. $2^2 = 1 \times 4$ gives $(a, b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Generating functions ( recurrence relations ) Find $a_n$ using Generating Functions : $a_n = -a_{n-1} + 2a_{n−2}$, $n\ge2$ and $a_0 = 1$, $a_1 = 2$.
Approach : So I will form a characteristic equation $ r^2 + r - 2 = 0$ whose roots are $r_1 = -2$, $r_2 = 1$.
So my general solution is $a_n = α_1r_1^n + α_2r_2^n$.
$a_n =... | If you really want to use generating functions to get $a_n$. In general with a recurrence relation with initial conditions $a_0$ and $a_1$ and \begin{equation}
a_n = r_1a_{n-1}+r_2a_{n-2}
\end{equation}
you can write the generating function as \begin{equation}
G(x) = \frac{-a_0-a_1x+a_0r_1x}{r_2x^2+r_1x-1}
\end{equatio... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Intersection of a hypersphere (4d) and a hyperplane I have a problem to solve and I really am in over my head here. Just to get me started, if I have a four dimensional plane and a 4 dimensional sphere
$ax + by + cz + dw = k$ and $x^2+y^2+z^2+w^2 = R$, and they intersect, am I expecting a three dimensional sphere.
In ... | Multiply the $4d$ sphere by $(a^2+b^2+c^2+d^2)$ & rearrange it to give
\begin{eqnarray*}
(ax+by+cz+dw)^2+(bx-ay+dz-cw)^2+(cx-dy-az+bw)^2+(dx+cy-bz-aw)^2=R^2(a^2+b^2+c^2+d^2)
\end{eqnarray*}
Now substitute the equation for the plane & let
\begin{eqnarray*}
X=bx-ay+dz-cw \\
Y=cx-dy-az+bw\\
Z=dx+cy-bz-aw
\end{eqnarray*}
a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can anyone help to proof convergence and find the sum of such series? May be correct my mistakes. $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$
I personally have such an idea: try to make geometric series like this
$$1-\frac{1}{\sqrt{10}}(1-\frac... | Write your sum like:
$$S = {S}_1 + {S}_2 + {S}_3$$
Where,
$${S}_1 = 1 + \frac{1}{\sqrt{10}\cdot 10} + \frac{1}{10^{3}} + ... = \frac{1}{1 - \frac{1}{10 \cdot\sqrt(10)}} = \frac{10\sqrt{10}}{10\sqrt{10} - 1}$$
$${S}_2 = -\frac{1}{\sqrt{10}} - \frac{1}{10^{2}} - ... = -\frac{1}{\sqrt{10}} \cdot{S}_1$$
$${S}_3 = -\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Number of values for a such that the following polynomial has only integral roots: $P(x) =X^3-X+a^2+a$? How do I solve this problem? I thought of solving it with Viete but I only have 2 relations from which I can find the roots.
| Let $(X-\alpha)(X-\beta)(X-\gamma) = X^3 - X + a^2 + a$ where $\alpha,\beta,\gamma \in \mathbb{Z}$. Then
$$\alpha + \beta + \gamma = 0, \qquad \alpha\beta + \alpha\gamma + \beta\gamma = -1, \qquad \alpha\beta\gamma = -a^2 -a.$$
From the first equation, $\gamma = -(\alpha + \beta)$. Then from the second equation
$$\alph... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the total number of partitions of $12$ having unequal positive parts. $P(k, n) = P(k - 1, n - 1) + P(k - n, n)$
Let $P^\star(k,n)$ denote the number of partitions of $k$ having exactly $n$ positive parts, all of which are unequal. For example, $P(8, 3)$ implies $8$ can be partitioned as $8 = 2 + 2 + 4,$ but $P^\st... | Your answer is correct. As a check, here is a list of the $15$ partitions of $12$ into unequal positive parts:
\begin{align*}
12 & = 12\\
& = 1 + 11\\
& = 2 + 10\\
& = 3 + 9\\
& = 4 + 8\\
& = 5 + 7\\
& = 1 + 2 + 9\\
& = 1 + 3 + 8\\
& = 1 + 4 + 7\\
& = 1 + 5 + 6\\
& = 2 + 3 + 7\\
& = 2 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve system in $a,b,c$ What's the best way to solve the system of equations ?:
$$
\left\{\begin{array}{rcccccr}
a & + & b & + & c & = & 1
\\
a^{2} & + & b^{2} & + & c^{2} & = & 7
\\
a^{3} & + & b^{3} & + & c^{3} & = & 13
\end{array}\right.
$$
If I square the first one I get $2\left(\,ab + bc + ac\,\right... | By using Newton's identities, we can find
that
$$s_2:=ab+bc+ca=\frac{1}{2}(a+b+c)^2-\frac{1}{2}(a^2+b^2+c^2)=-3$$
and
$$s_3:=abc=\frac{1}{3}(a^3+b^3+c^3)-\frac{1}{3}(a+b+c)^3+(ab+bc+ca)(a+b+c)=1.$$
Then we obtain $(a,b,c)$ by solving the polynomial equation
$$x^3-s_1x^2+s_2x-s_3=x^3-x^2-3x-1=(x+1)(x^2-2x-1)\\=(x+1)... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Compute $\sum_{0}^{\infty}(-1)^k \frac{(k + 1) ^ 2}{k!}$ Compute
$$\sum_{0}^{\infty}(-1)^k \frac{(k + 1) ^ 2}{k!}.$$
Today I have participated in Olympiad which had 8 different tasks. This was one of them.
I've started by substituting concrete values and looking for a pattern:
$\sum_{0}^{\infty}(-1)^k \frac{(k + 1) ... | HINT:
As $(k+1)^2=k(k-1)+3k+1,$
For $k\ge2,$
$$\dfrac{(k+1)^2}{k!}=\dfrac1{(k-2)!}+3\cdot\dfrac1{(k-1)!}+\dfrac1{k!}$$
$$\dfrac{(k+1)^2}{k!}x^k=x^2\cdot\dfrac{x^{k-2}}{(k-2)!}+3x\cdot\dfrac{x^{k-1}}{(k-1)!}+\dfrac{x^k}{k!}$$
For $k=0,$ $$\dfrac{(k+1)^2}{k!}x^k=1$$
For $k=1,$ $$\dfrac{(k+1)^2}{k!}x^k=4x=3x\cdot\dfrac1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving Trigonometric Equality I have this trigonometric equality to prove:
$$\frac{\cos x}{1-\tan x}-\frac{\sin x}{1+\tan x}=\frac{\cos x}{2\cos^2x-1}$$
I started with the left hand side, reducing the fractions to common denominator and got this:
$$\frac{\cos x+\cos x\tan x-\sin x+\sin x\tan x}{1-\tan^2x}\\=\frac{\cos... | $$\frac{c^2}{c-s}-\frac{cs}{c+s}=\frac{\cos x}{2\cos^2x-1}$$
$$\frac{c^3+sc^2-c^2s+cs^2}{c^2-s^2} =\frac{\cos x}{2\cos^2x-1}$$
$$\frac{ c }{c^2-s^2}=\frac{\cos x}{2\cos^2x-1}$$
$$\frac{ c }{2c^2-1}=\frac{\cos x}{2\cos^2x-1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Is the transformation possible? I came across this problem which asks to transform $x^2-x-2$ to $x^2-x-1$ ,if possible, using the following rules:
Given a quadratic equation $ax^2+bx+c$ you can :
1)Interchange $a$ and $c$
2)Replace $x$ by $x+t$ where $t$ is a real number.
My approach:
I wrote the equations as
${(x-\fr... | We start with a quadratic $ax^2+bx+c$ and use one of the steps to transform it to $a'x^2+b'x+c'$.
In the first case, we have $a'=c,b'=b,c'=a$, in the second case, we have $a'=a, b'=b+2at, c'=c+bt+t^2$.
Note that in the first case, we have $b'^2-4a'c'=b^2-4ac$ and in the second case $b'^2-4a'c'=b^2+4abt+4a^2t^2-4a(c+bt+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve $p^{2a+1}+p^a+1=b^2$ in integers? For prime number $p$ and positive integers $a$ and $b$ solve the following equation
$$p^{2a+1}+p^a+1=b^2$$
Can you give me a hint for starting this problem?
| Hint
$$p^a(p^{a+1}+1)=b^2-1=(b-1)(b+1)$$
Hint 2 $gcd(b-1,b+1)=1$ or $2$. If $p \neq 2$ then $p^a$ must divide either $b-1$ or $b+1$.
Hint 3
If $b+1=kp^{a}$ then $b-1=\frac{p^{a+1}+1}{k}$. Then
$$kp^a-\frac{p^{a+1}+1}{k}=2 \\
k^2p^a-p^{a+1}-1=2k \\
k^2p^a-p^{a+1}=2k+1 \\
$$
This means that $p |2k+1$.
Also, $b+1=kp^... | {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate product $\prod_{k=0}^{n-1}\left (1+\frac{1}{2^{2^k}}\right)$? How can I calculate the following product of series? $$ \prod_{k=0}^{n-1}\left (1+\frac{1}{2^{2^k}}\right) $$....Can I take a geometric series and compare it with that?
| As a hint, consider $n = 3$:
$\displaystyle\prod_{k = 0}^{2}\left(1+\dfrac{1}{2^{2^k}}\right) = \left(1+\dfrac{1}{2^1}\right)\left(1+\dfrac{1}{2^2}\right)\left(1+\dfrac{1}{2^4}\right)$
$= 1 + \dfrac{1}{2^1} + \dfrac{1}{2^2} + \dfrac{1}{2^1} \cdot \dfrac{1}{2^2} + \dfrac{1}{2^4} + \dfrac{1}{2^1} \cdot \dfrac{1}{2^4} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the closed form of $\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b\sin^2x+c\cos^2x}\mathrm dx=f(a,b,c)$ Last question of this form
I am very curious to what is the closed form of:
Assume where $a,b,c > 0$
$$\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b\sin^2x+c\cos^2x}\mathrm dx=f(a,b,c)\tag1$$
$$b\sin^2x+c\cos^2x$$... | We have
$$ f(a,b,c) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan(a \tan^2\theta)}{b \sin^2\theta + c\cos\theta^2} \, d\theta = \frac{\pi}{\sqrt{bc}} \left( \arctan\left(1 + \sqrt{\frac{2ac}{b}}\right) - \frac{\pi}{4} \right). $$
Here is a sketch of computation.
Step 1. Reduction of the problem (not essential)
Consider t... | {
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shortcut method for calculating a determinant \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c& a+b \end{vmatrix} = 4abc
Is there any shortcut method to compute this determinant without breaking by definition. Need some help please.
| $$\begin{vmatrix}
b+c & a & a \\ b & c+a & b \\ c & c& a+b
\end{vmatrix} \xrightarrow{-R_1+R_2} \begin{vmatrix}
b+c & a & a \\ -c & c & -a+b \\ c & c& a+b
\end{vmatrix} \xrightarrow{-R_2+R_3} \begin{vmatrix}
b+c & a & a \\ -c & c & -a+b \\ 2c & 0 & 2a
\end{vmatrix} \xrightarrow{-0.5R_3+R_1} \begin{vmatrix}
b & a & 0... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Radius of converge of Laurent series for $\frac{1}{\sin z}$ I want to show that the radius of convergence of Laurent series for $\frac{1}{\sin z}$ is $\pi$
I showed that:
\begin{align}
\frac 1 {\sin z} & = \frac 1 {z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots } = \frac 1 {z\big(1 - (\frac{z^2}{3!} - \frac{z^4}{... | After writing my comment, I realize that it is way simpler...apparently:
The function $\;\cfrac1{\sin z}\;$ is analytic at $\;\left\{\,z\in\Bbb C\;/\;0<|z|<\pi\,\right\}\;$ and thus the radius of convergence of (the non principal part of) the Laurent series of the function around zero is at least $\;\pi\;$ . But as alr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Arclength, finding definite integral Find the length of the segment between $x=5$ and $x=2$ for the following curve.
$$f(x)= \frac{1}{2}x^2-\frac{1}{4}\log\left(x\right)$$
My working so far.
$$f'\left(x\right)=x-\frac{1}{4x}.$$
Then
$$L = \int _2^5\:\sqrt{1+\left(f'\left(x\right)\right)^2}\,dx
=\int _2^5\:\sqrt{x^2+\fr... | $$L = \int _2^5\:\sqrt{1+\left(f'\left(x\right)^2\right)}
\\
\\=\int _2^5\:\sqrt{1+\left(x-\frac{1}{4x}\right)^2}\\
\\=\int _2^5\:\sqrt{1+x^2-\dfrac{1}{2}+\dfrac{1}{(4x)^2}}dx\\
=\int _2^5\:\sqrt{x^2+\dfrac{1}{2}+\dfrac{1}{(4x)^2}}dx\\
\\=\int _2^5\:\sqrt{\left(x+\frac{1}{4x}\right)^2}\\
=\int _2^5\:|x+\frac{1}{4x}|dx ... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the product is never a perfect square
Prove that for nonnegative integers $x_1,\ldots,x_{2011}$ and $y_1,\ldots,y_{2011}$ the product $$(2x_1^2+3y_1^2)(2x_2^2+3y_2^2) \cdots (2x_{2011}^2+3y_{2011}^2)$$ is never a positive perfect square.
I thought about generalizing this question to any odd subscript $n$ i... | Another viewpoint is by using the composition of binary quadratic forms, to reduce the case $ n $ odd to the case $ n = 1 $, in a `systematic' way. We have the following identities writing a product
$$
(2a^2 + 3b^2)(2c^2 + 3d^2) = (3 b c + 2 a d)^2 + 6 (-a c + b d)^2
$$
and
$$
(2a^2+3b^2)(c^2 + 6y^2) = 2 (3 b c + a ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show which of $6-2\sqrt{3}$ and $3\sqrt{2}-2$ is greater without using calculator How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)
Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the fi... | $6-2√3 \approx2(1.732) = 6 - 3.464 = 2.536$
$3√2-2 \approx 3(1.414) - 2 = 4.242 - 2 = 2.242$
This implies that $\dfrac 52$ is between the two quantities:
\begin{align}
\dfrac{49}{4} &> 12 \\
\dfrac 72 &> 2\sqrt 3 \\
\dfrac{12}{2} &> 2\sqrt 3 + \dfrac 52 \\
6 - 2\sqrt 3 &> \dfrac 52
\end{align}
and
\begin{al... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find all real coefficient polynomial such $f(x)|f(x^2-2)$
Find all real coefficient polynomial with the degree is $3$,and such
$$f(x)|f(x^2-2),\forall x\in Z $$
I try: let $f(x)=ax^3+bx^2+cx+d$,then $$f(x^2-2)=ax^6-6ax^4+12ax^2-8a+bx^4-4bx^2+4b+cx^2-2c+d$$,then such
$$(ax^3+bx^2+cx+d)|ax^6-6ax^4+12ax^2-8a+bx^4-4bx... | Hint: Write $$f(x^2-2)=g(x)f(x)$$ If $x$ is a root of $f$ then $x^2-2$ is a root too. Then $(x^2-2)^2-2$ is a root of $f$ and etc. But $f$ cannot have infinitely many roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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"answer_id": 1
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What is the probability that 2 or more numbers out of 4 are among the drawn lottery numbers (6 out of 49 without replacement)? I have the following combinatoric problem to solve; I have searched and couldn't find an answer already existing.
In the main lottery in Germany, 6 numbers are to be drawn out of 49 {1,2,…49} w... | We can consider three cases:
*
*Draw four numbers out of six. All four balls must belong to the six chosen balls, so the probability of this happening equals:
$$\frac{6}{49} \cdot \frac{5}{48} \cdot \frac{4}{47} \cdot \frac{3}{46}$$
*Draw three numbers out of six. One way to do this, is by first choosing three ball... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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A polynomial is divisible with another one For what $m$ and $n$ is the polynomial $2X^{19}+X^{13}+mX^{11}+X^8+2X^6+nX^2+2$ divisible by $X^4+X^3+X^2+X+1$.
I tried to find the real solutions for g but couldn't
| HINT:
$$x^5-1=(x-1)(x^4+x^3+x^2+x+1)=0$$
$$\implies x^{19}=x^4,x^{13}=x^3$$
| {
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"answer_id": 0
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Let $a^4+b^4+c^4=3$. Prove that $a^2+b^2+c^2\geq a^2b+b^2c+c^2a$ Let $a^4+b^4+c^4=3$. Prove that $a^2+b^2+c^2\geq a^2b+b^2c+c^2a$
My proof :D
We have the inequality
$\sum_{cyc}^{ }a^2.\sum_{cyc}^{ }a\geq \sum_{cyc}^{ }a^2b $
which is equivalent to $\sum_{cyc}^{ }b(a-b)^2$ (true)(1)
In another side, by AM-GM, we have: $... | It's enough to prove our inequality for non-negative variables.
We'll prove a stronger inequality:
$$a^2b+b^2c+c^2a\leq3\left(\sqrt{\frac{a^2+b^2+c^2}{3}}\right)^3,$$
which solves our problem.
Indeed, by C-S $$a^2+b^2+c^2=\sqrt{(a^2+b^2+c^2)^2}\leq\sqrt{(1+1+1)(a^4+b^4+c^4)}=3.$$
Thus,
$$a^2b+b^2c+c^2a\leq3\left(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find this limit. Compute the value of the limit :
$$
\lim_{x\to\infty}{\frac{1-\cos x\cos2x\cos3x}{\sin^2x}}
$$
I've tried simplifying the expression to
$$
\lim_{x\to\infty}\frac{-8\cos^6x+10\cos^4x-3\cos^2x+1}{\sin^2x}
$$
But I don't know what to do after this.
| Letting
$c = \cos(x)$,
$\cos x\cos2x\cos3x
=c(2c^2-1)(4c^3-3c)
=c^2(2c^2-1)(4c^2-3)
$
so
$\dfrac{1-\cos x\cos2x\cos3x}{\sin^2x}
=\dfrac{1-c^2(2c^2-1)(4c^2-3)}{1-c^2}
=8 c^4 - 2 c^2 + 1
$
(according to Wolfy).
Putting
$c^2 = 1-s^2$
(s = sin),
this becomes
$8(1-s^2)^2-2(1-s^2)+1
=8(1-2s^2+s^4)-2+2s^2+1
=8s^4-14s^2+7
$.
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Calculate the sum of the series $\sum_{1\leq aCalculate the sum of the series:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c}$$
My attempt:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c} = \sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}$$
Is it... | HINT: How about rewriting the sum as
$$S=\sum_{1\leq a<b<c\\a,b,c\in\mathbb{N}}\frac{1}{2^a 3^b 5^c}
=\sum_{a=1}^{\infty}\sum_{b=a+1}^{\infty}\sum_{c=b+1}^{\infty}\frac{1}{2^a 3^b 5^c}=\sum_{a=1}^{\infty}\frac{1}{2^a}\sum_{b=a+1}^{\infty}\frac{1}{3^b}\sum_{c=b+1}^{\infty}\frac{1}{5^c}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
prove that : if $a, b \in \mathbb{R}^+$ : then : $a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$ prove that : if $a, b \in \mathbb{R}^+$ : then :
$$a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$$
$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2b+b^2a-a^2$$
$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2+b^3a-b^2$$
now what ?
| Let $ab=x$, $a^2+b^2=y$, then $0\leq 2x\leq y$ and we want
$$x^2(y-2)\geq y(x-1)$$
$$x^2y-2x^2\geq xy-y$$
$$y\geq\frac{2x^2}{x^2-x+1}$$
It is equivalent to show
$$2x\geq\frac{2x^2}{x^2-x+1}$$
$$x^2-x+1\geq x$$
$$(x-1)^2\geq0$$
Equality holds iff $a=b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$ I came across an interesting equation with two variables $x,y\in\mathbb{R}$, $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$
This, once expanded, can be simplified to $3x^2 - 3xy + 3y^2 - 3y + 1=0.$
How can one proceed to solve it algebraically? The solution according to ... | You can use the Cauchy-Schwarz Inequality:
$$(1+1+1)\big(x^2+(1-y)^2+(y-x)^2\big)\geq \big(1\cdot x+1\cdot(1-y)+1\cdot(y-x)\big)^2=1^2=1\,.$$
Thus,
$$x^2+(1-y)^2+(y-x)^2\geq\frac{1}{3}\,.$$
The inequality becomes an equality if and only if $$\frac{x}{1}=\frac{1-y}{1}=\frac{y-x}{1}\,,$$ or equivalently $$(x,y)=\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
number of divisors of $2^23^35^57^411^3$ which are is in the form of $6k+1, k\geq 0$ and $k\in \mathbb{Z}$
Total number of divisors of $2^23^35^57^411^3$ which are is in the form of $6k+1, k\geq 0$ and $k\in \mathbb{Z}$
$\bf{Attempt}$ writting $1,3,3^2,3^3$ as $6k+1$ or $6k+3$
same way $1,5,5^2,5^3,5^4,5^5$ as $6k+... | Well, we can get $1 = 6k+1$ for $k=0$ off the bat, so that's nice. We can also ignore the $2$s and $3$s (we may do so by arguing that no divisor of $2^{2}3^{3}5^{5}7^{4}11^{3}$ with $2$ or $3$ as a factor will be congruent to $1$ mod $6$). Fix $a \in \mathbb{N}$. Since $7 = 6k+1$ for $k=1$, we clearly have $7^{a} \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Integrate Airy function from 0 to $\infty$ From the integral representation of Airy function
$$\mathrm{Ai}(x)=\int_{-\infty}^{\infty} \frac{\mathrm{d} \tau}{2\pi} \exp(-\mathrm{i}\tau x)\exp(-\mathrm{i}\frac{\tau^3}{3}),$$
It is easy to see that $\int_{-\infty}^{\infty} \mathrm{d} x\mathrm{Ai}(x) =1$. However, I am won... | EDIT:
After posting my answer, I realized that the same approach can be used to find the Mellin transform of $\operatorname{Ai}(x)$. So I decided to modify my answer.
I don't know if this is the approach discussed in the reference, but the Airy function $\operatorname{Ai}(x)$ can be expressed in terms of the modified... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.