Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Transformation of general equation of straight line to normal form Consider the general form of equation of straight line.
$$ax + by + c = 0$$
$$\Rightarrow ax + by = -c$$
$$\Rightarrow \frac{ax}{-c} + \frac{by}{-c} = 1$$
$$\Rightarrow \frac{x}{\frac {-c}{a}} + \frac{y}{\frac {-c}{b}} =1 $$
this is equation of line in ... | The general equation $ax+by+c=0$ of a straight line in $\mathbb R^2$ can be rewrtten as $\mathbf n\cdot\mathbf x=-c$, with $\mathbf n\ne0$. This form of the equation tells us that a line can be characterized as the set of points that have the same dot product with some fixed vector $\mathbf n$. If we normalize $\mathbf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{ n \to \infty }(1-\tan^2\frac{x}{2})(1-\tan^2\frac{x}{4})(1-\tan^2\frac{x}{8})...(1-\tan^2\frac{x}{2^m})=?$ Find the limit :
$$\lim_{ n \to \infty }(1-\tan^2\frac{x}{2})(1-\tan^2\frac{x}{4})(1-\tan^2\frac{x}{8})...(1-\tan^2\frac{x}{2^n})=?$$
My try :
$$1-\tan^2 y = \frac{2\tan y }{\tan(2y)}$$
$$\lim_{ n ... | Note that the numbers in the deonominator and numerator cancel, leaving only $$\lim_{n \to \infty} 2^{n} \frac{\tan \frac{x}{2^n}}{\tan x}$$
In your original equation. However, as $\lim\limits_{n \to \infty} \frac{x}{2^n}=0$, we have that $$\lim_{n \to \infty} 2^n \tan \frac{x}{2^n}=\lim_{n \to \infty}x \times \dfrac{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
| Observe that $7^2=49<2\cdot 5^2$. In this case,
$$
7^{55}=7\cdot 7^{54}=7\cdot(7^2)^{27}<7\cdot (2\cdot 5^2)^{27}=7\cdot 2^{27}5^{54}.
$$
Observe that $2^3<10=2\cdot 5$. In this case,
$$
7\cdot 2^{27}5^{54}=7\cdot (2^3)^95^{54}<7\cdot(2\cdot 5)^95^{54}=7\cdot 2^9\cdot 5^{63}
$$
Using that $2^3<10=2\cdot 5$ again, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 4
} |
Find the limit : $\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$ Find the limit: Without the use of the L'Hôspital's Rule
$$\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$$
My try:
$u=x-1$
Now:
$$\lim_{ x \to 1}\frac{\sqrt[n]{(u+1)^n-1}}{\sqrt[n]{n(u+1)}-... | Let us assume $n > 1$. We must have $x \to 1^{+}$ to ensure that the roots are well defined for all $n > 1$. We can proceed as follows
\begin{align}
L &= \lim_{x \to 1^{+}}\frac{\sqrt[n]{x^{n} - 1}}{\sqrt[n]{nx} - \sqrt[n]{n} - \sqrt[n]{nx - n}}\notag\\
&= \frac{1}{\sqrt[n]{n}}\lim_{x \to 1^{+}}\frac{\sqrt[n]{x^{n} - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\arctan(1/3)$ How can I calculate $\arctan\left({1\over 3}\right)$ in terms of $\pi$ ? I know that $\tan^2(\frac{\pi}{6})= {1\over3}$ but don't know if that helps in any way.
| For the evaluation of $\tan^{-1}(x)$, you can also use Padé approximants such as
$$\frac{ x+\frac{4 x^3}{15}} { 1+\frac{3 x^2}{5}}\tag 1$$
$$\frac{x+\frac{11 x^3}{21} } {1+\frac{6 x^2}{7}+\frac{3 x^4}{35} } \tag 2$$
$$\frac{x+\frac{7 x^3}{9}+\frac{64 x^5}{945}} {1+\frac{10 x^2}{9}+\frac{5 x^4}{21} } \tag 3$$
$$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle what is the value of ab? If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle, find the value of $ab$
| I'd like to propose an exotic solution using complex numbers. Let the three vertices be represented on the Argand diagram by
$$z_0 = 0 + 0i = 0$$
$$z_1 = a + 11i$$
$$z_2 = b + 37i$$
Then we know that $z_1-z_0 = z_1$ and $z_2-z_0 = z_2$ are $\frac{\pi}{3}$ apart. That is
$$z_1 = z_2e^{\pm i\frac{\pi}{3}}$$
$$a + 11i = (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving for a matrix $$
\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}X = \begin{bmatrix}3&2&1\\6&5&4\\9&8&7\end{bmatrix}.
$$
I know how to do this normally, but I'm confused with this case. It seems like an inverse does not exist for the matrices, and the second is the first matrix in reverse.
| Recall that, when we multiply matrices $A$ and $B$, each column of $AB$ is a linear combination of the columns of $A$, with weights given by entries in the corresponding column of B.
In this case, you want the first column of $X$ to be something that says, "ignore the first two columns of $A$, and take 1 times the thir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show that $a+b+c=0$ implies that $32(a^4+b^4+c^4)$ is a perfect square. There are given integers $a, b, c$ satysfaying $a+b+c=0$. Show that $32(a^4+b^4+c^4)$ is a perfect square.
EDIT: I found solution by symmetric polynomials, which is posted below.
| It suffices to show that $2(a^4+b^4+c^4)$ is a perfect square.
\begin{align}
& 2(a^4+b^4+c^4) \\
=& 2((b+c)^4+b^4+c^4) \\
=& 2(2b^4+4b^3c+6b^2c^2+4bc^3+2c^4) \\
=& 4(b^4+\bbox[2px, border:1px solid]{2b^3c}+\bbox[2px, border:1px dashed]{3b^2c^2}+2bc^3+c^4) \\
=& 4((b^4+\bbox[2px, border:1px solid]{b^3c}+\bbox[2px, borde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Find matrix $B$ such that $BA=4A$ Find the unique $B \in \mathbb{R}^{3 \times 3}$ such that for every $A \in \mathbb{R}^{3 \times 3}$ we have
i) $BA=4A$
ii) The 1, 2 and 3 rows of $BA$ are the 3,2 and 1 rows of A.
This problem is in the section where they define matrix multiplication. My only idea was to set up a gia... | Taking them to be separate questions,
You should start by looking at what $4A$ is equal to as an arbitrary matrix. Then, seeing what matrix can a multiply by in order to get this this product? In the second case, the same reasoning follows. See below. Let $A$ be the matrix with a standard representation.
*
*$4A = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Closed form for the series $\sum_{k=1}^\infty (-1)^k \ln \left( \tanh \frac{\pi k x}{2} \right)$
Is there a closed form for: $$f(x)=\sum_{k=1}^\infty (-1)^k \ln \left( \tanh \frac{\pi k x}{2} \right)=2\sum_{n=0}^\infty \frac{1}{2n+1}\frac{1}{e^{\pi (2n+1) x}+1}$$
This sum originated from a recent question, where we h... | Let’s use $~\displaystyle\prod\limits_{k=1}^\infty (1+z^k)(1-z^{2k-1}) =1~$ . $\enspace$ (It's explained in a note below.)
For $~z:=q^2~$ and $~q:=e^{-\pi x}~$ with $~x>0~$ we get
$\displaystyle e^{f(x)} = \prod\limits_{k=1}^\infty\frac{\tanh(k\pi x)}{\tanh((k-\frac{1}{2})\pi x)} = \prod\limits_{k=1}^\infty\frac{ \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Solving: $(b+c)^2=2011+bc$
Solve $$(b+c)^2=2011+bc$$ for integers $b$ and $c$.
My tiny thoughts:
$(b+c)^2=2011+bc\implies b^2+c^2+bc-2011=0\implies b^2+bc+c^2-2011=0$
Solving in $b$ as Quadratic.$$\implies b=\frac{-c\pm \sqrt{8044-3c^2}} {2}.$$
So $8044-3c^2=k^2$, as $b$ and $c$ are integers. We also have inequalitie... | We can assume that $b\geq0$ and by your starting reasoning we ca get $b\in\{10,39,49\}$,
which gives all 12 solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Quadratic equation find all the real values of $x$ Find all real values of $x$ such that
$\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$
I tried sq both sides by taking 1 in RHS but it didn't worked out well...
| HINT: after squaring one times we get
$$2\sqrt{x-\frac{1}{x}}\sqrt{1-\frac{1}{x}}=x^2-x+\frac{2}{x}-1$$
can you finish this?
squaring this one more times we get
$$4\left(x-\frac{1}{x}\right)\left(1-\frac{1}{x}\right)=\left(x^2-x+\frac{2}{x}-1\right)^2$$
expanding the left Hand side we obtain $$4\,x-4-4\,{x}^{-1}+4\,{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Inequality for positive real numbers less than $1$: $8(abcd+1)>(a+1)(b+1)(c+1)(d+1)$ If $a,b,c,d$ are positive real numbers, each less than 1, prove that the following inequality holds:
$$8(abcd+1)>(a+1)(b+1)(c+1)(d+1).$$
I tried using $\text{AM} > \text{GM}$, but I could not prove it.
| Given that $0<a<1$, $0<b<1$, $0<c<1$, $0<d<1$
$\therefore$ $-1<a-1<0$ and $-1<b-1<0$
Now
\begin{align*}
&(a-1)(b-1)>0 \\
\Rightarrow& ab - (a+b) +1 >0 \\
\Rightarrow& (ab+1) > a+b \\
\Rightarrow& (ab+1)+(ab+1) > (ab+1)+(a+b) \\
\Rightarrow& 2(ab+1) > (ab+1)+(a+b) \\
\Rightarrow& 2(ab+1) > (a+1)(b+1) \tag{I}
\end{align*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$ Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$.
Find $(a+b+c)$.
I computed the whole product ;If $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to p... | Since scaling by a constant
leaves the product unchanged,
we can assume that
$a=1$.
This becomes
$(1+b+c)(1+1/b+1/c)
=16$.
Looking for a special solution,
assume $b = c$.
This becomes
$16
=(1+2b)(1+2/b)
=1+2b+2/b+4
$
or
$11
=2b+2/b
$
or
$2b^2-11b+2 = 0$.
Solving this,
$b
=\dfrac{11\pm \sqrt{11^2-16}}{4}
=\dfrac{11\pm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Multiple choice question of indefinite integral, $\int \frac{x + 9}{x^3 + 9x} dx$. If $\int \frac{x + 9}{x^3 + 9x} dx = k\arctan(mx) + n\ln (x) + p \ln (x^2 + 9) + c$, then $(m+n)/(k+p) = $
(A) 6
(B) -8
(C) -3
(D) 4
I tried solving it by differentiating the R.H.S. but couldn't arrive at the answer.
| \begin{align}
\int \frac{x + 9}{x^3 + 9x} dx &= \int \frac{x}{(x^2 + 9)x} dx + \int \frac{9}{(x^2 + 9)x} dx
\\
\\
&= \int \frac{dx}{x^2 + 9} + \int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 9}\right) dx
\\
\\
&= \int \frac{dx}{x^2 + 9} + \int \left(\frac{1}{x} + \frac{-x}{x^2 + 9}\right) dx
\\
\\
&= \frac{1}{9}\int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
Prove the following:
$\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
How do you go about proving this?
| \begin{align}\sqrt{2-\sqrt{3}} &=\frac{\sqrt{2}\sqrt{2-\sqrt{3}}}{\sqrt{2}}\\
&=\frac {\sqrt { 4-2\sqrt{3}}}{\sqrt{2}} \\
&=\frac {\sqrt{1-2\sqrt{3} +{ \left( \sqrt{3}\right)}^{2} } }{ \sqrt {2} } \\
&=\frac{\sqrt { { \left( 1-\sqrt{3}\right)}^{2}}}{ \sqrt{2}}\\
&=\frac{\sqrt {3} -1 }{ \sqrt {2} }\\
& = \frac{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$ If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$.
Find the value of $2b + \dfrac {c}{a}$.
My Attempt:
$$\sin A+\sin^2 A=1$$
$$\sin A + 1 - \cos^2 A=1$$
$$\sin A=\cos^2 A$$
Now,
$$a(\cos^2 A)^{6}+b(\cos^2 A)^{4}+c(\cos^2 A)^{3}=1$... | $$\sin A= \cos ^2A→\sin^2A=\cos^4A→1-\cos^2A=\cos^4A\\
(\cos^2A)^2+(\cos^2A)-1=0→\cos^2A=\frac{-1+ \sqrt{5}}{2}$$
So,
$$a\left(\frac{-1+ \sqrt{5}}{2}\right)^6+b\left(\frac{-1+ \sqrt{5}}{2}\right)^4+c\left(\frac{-1+ \sqrt{5}}{2}\right)^3=1\\
a(9-4\sqrt{5})+b\left(\frac{7-3\sqrt{5}}{2}\right)+c(\sqrt{5}-2)=1\\
(9a+7b/2-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find radius of convergence and interval of converges of the series; $\displaystyle\sum_{n=0}^{\infty}\frac{1}{(-3)^{2+n}(n^2+1)}(4x-12)^n$ $\displaystyle\sum_{n=0}^{\infty}\frac{1}{(-3)^{2+n}(n^2+1)}(4x-12)^n$
My try:$$\displaystyle\lim_{n\to\infty}\left(\frac{1}{(-3)^{2+n}(n^2+1)}(4x-12)^n\right)^{\frac{1}{n}}\leq1$$ ... | I personally prefer the ratio test. If a series converges, then:
$$\lim_{k\to \infty}\left|\frac{a_{k+1}}{a_k}\right|<1$$
Substituting gives:
$$\lim_{k\to \infty}\left|\frac{\left(\frac{(4x-12)^{k+1}}{(-3)^{3+k}((k+1)^2+1)}\right)}{\left(\frac{(4x-12)^{k}}{(-3)^{2+k}(k^2+1)}\right)}\right|<1$$
$$\lim_{k\to \infty}\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can this symmetric matrix be an orthogonal matrix? So the question is to prove whether this matrix M is orthogonal?
$$
M = \begin{bmatrix}
a & k & k \\
k & a & k \\
k & k & a \\
\end{bmatrix}
$$
My attempt to find the inverse of M :
$$det(M)=(a+2k)(a-k)(a-k)$$
The inverse of $M$ ... | Since $M$ is symmetric, $M=M^T$, so to check orthogonality, we compute $MM^T=M^2$ and get
$$
\begin{bmatrix}
a^2+2k^2&2ak+k^2&2ak+k^2\\
2ak+k^2&a^2+2k^2&2ak+k^2\\
2ak+k^2&2ak+k^2&a^2+2k^2
\end{bmatrix}
$$
Therefore, you need $2ak+k^2=0$ and $a^2+2k^2=1$. Factoring the first equation, we have $k(2a+k)=0$. Therefore, e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving $\frac{x^2 + k^2 - 2xk}{x^2 + c^2 - 2xc} = e^{-2yg}\frac{k^2}{c^2}$ $$\frac{x^2 + k^2 - 2xk}{x^2 + c^2 - 2xc} = e^{-2yg}\frac{k^2}{c^2}$$
I am attempting to solve for $x$. I've been at it for a while and have reached to above equation, but now it seems to get quite messy. Any help?
Important:
$k = \frac{a+g}{2E... | It helps to note that everything is pretty much a constant, hence we may reduce as follows:
$$\frac{x^2-2kx+k^2}{x^2-2cx+c^2}=P$$
where $P=e^{-2yg}\frac{k^2}{c^2}$. Then notice some perfect squares:
$$x^2-2kx+k^2=(x-k)^2\\x^2-2cx+c^2=(x-c)^2$$
Thus, the problem further reduces:
$$\frac{x-k}{x-c}=\pm\sqrt P$$
Multiply ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2169415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is $f_n(z) = \dfrac {2nz - 1} {z + n^2}$ uniformly convergent on the disc $|z| <1$? The pointwise limit of $f_n(z) = \dfrac {2nz - 1} {z + n^2}$ is $0$.
By the triangle inequality,
$\Bigg|\dfrac {2nz - 1} {z + n^2}\Bigg| \le \Bigg|\dfrac {2nz} {z + n^2}\Bigg| + \Bigg|\dfrac {1} {z + n^2}\Bigg|$
On the disc $|z| <1$, $\... | No, $|1/(z+n^2)| \le 1/n^2$ fails. Try $z=-1/2$ for example. Instead note that for $n>1,$
$$\left |\frac{2nz-1}{z+n^2}\right | \le \frac{|2nz|+1}{n^2 - |z|} \le \frac{2n+1}{n^2 - 1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine: $S = \frac{2^2}{2}{n \choose 1} + \frac{2^3}{3}{n \choose 2} + \frac{2^4}{4}{n \choose 3} + \cdots + \frac{2^{n+1}}{n+1}{n \choose n}$ We are given two hints: consider $(n+1)S$; and use the Binomial Theorem. But we are not to use calculus.
My consideration of $(n+1)S$ goes like this:
\begin{align*}
\sum\... | Note that
$$\sum_{k=1}^n2^{k+1}{n+1\choose k+1}=\sum_{m=2}^{n+1}2^m{n+1\choose m}$$
So
$$\sum_{k=1}^n\frac{2^{k+1}}{k+1}{n\choose k}=\frac{1}{n+1}\sum_{m=2}^{n+1}2^m{n+1\choose m}=\frac{1}{n+1}\left((1+2)^{n+1}-1-2(n+1)\right)=\frac{3^{n+1}-3-2n}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
why is $2^{\log^{2}\frac{1}{2}} = 1/2$? I have no understanding of $\log^2n$, wouldn't $2^{\log^{2}\frac{1}{2}} = 1/4$ since $2^{\log\frac{1}{2}} = 1/2$?
$\log$ here has the base of $2$.
| $\textbf{Remark}$: Note that we have that $(a^b)^2 = a^{2b}$ and $a^{(b^2)} = a^{b^2}$, but in general we have that
$$a^{b^2} \neq a^{2b}$$ (unless $b^2 = 2b$, which is only true for $b = 2$ and $b = 0$).
(Consider for example $(2^3)^2 = 2^(3 \cdot 2) = 2^6 = 64$, whereas $2^{(3^2)} = 2^9 = 512$).
In your case, we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to show that $\sqrt{3 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{3 + \sqrt{2}})$ Is there a simple way to show that $\sqrt{3 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{3 + \sqrt{2}})$ (assuming this is true), using Galois theory? I have tried to show this by computational means but the calculations are horrible.
| Let $\,K\,$ denote $\,\mathbb Q(\sqrt{3+\sqrt2})\,$ for convenience, then observe that
$$x=\sqrt{3+\sqrt2}$$
$$\Rightarrow\quad x^2-3=\sqrt2\qquad\ \ \ $$
$$\Rightarrow\quad x^4-6x^2+7=0\quad$$
$$\Rightarrow\quad[\,K:\mathbb Q\,]=4\qquad\ \ \ $$
First, $\,(\sqrt{3+\sqrt2})^2-3=\sqrt2\in K$
Now if $\,\sqrt{3-\sqrt2}\in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Finding $a,b,c$ using the equation provided Provided with the equation
$$\sin^2 (n+1)x - \sin^2 (n-1)x = \sin2nx \sin2x$$
Find the values of $a,b,c$ if
$$ \sin^2 3x - \sin^2 x = 8\cos^2 x (a\cos^4 x + b\cos^2 x + c)$$
What I have tried
$(1 - \cos^2 3x)- (1- \cos^2 x)=8\cos^2 x (a\cos^4 x + b\cos^2 x + c)$
Expanding out... | Friend, your answer is wrong,
Correct answer is $a=-2$, $b=3$ and $c=-1$
For a quick verification, put $x=0$ ;
LHS = $0$
thus RHS must be $0$, i.e. $a+b+c=0$. Which isn't true in your case.
Proper proof using given equation :
$\sin^2(n+1)x−\sin^2(n−1)x=\sin2nx\sin2x$
Put $n=2$
$\sin^23x−\sin^2x= \sin4x\sin2x$
$= 2\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is there another transformation matrix for the bases of the image and of the preimage of this mapping? There is the following Matrix:
\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix}
At a point it is needed to calculate the determinant of the matrix. In the official solution it is written:
$det\begin{pmatrix... | Subtracting the first colum from the second and third columns we get
$$
\det\begin{pmatrix}1&1&1\\ a& b& c\\a^2& b^2& c^2\end{pmatrix}=\det\begin{pmatrix}1&0&0\\a&b-a&c-a\\ a^2&b^2-a^2&c^2-a^2\end{pmatrix}
$$
It follows that
\begin{eqnarray}
\det\begin{pmatrix}1&1&1\\ a& b& c\\a^2& b^2& c^2\end{pmatrix}&=&\det\begin{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
My work so far:
1) $$y =\frac{2x+9}{\sqrt{-x^2+4x+12}+\sqrt{-x^2+2x+3}} $$
2) I used a derivative and found the answer ($y=\sqrt3$ at $x=0$). Is there any other way?
| Rewriting the function that way only complicates things.
First let's determine the domain:
\begin{cases}
-x^2+4x+12\ge0\\
-x^2+2x+3\ge0
\end{cases}
reduces to $-1\le x\le3$.
The derivative is
$$
y'=
\frac{-x+2}{\sqrt{-x^2+4x+12}}-
\frac{-x+1}{\sqrt{-x^2+2x+3}}
$$
(which is undefined at $-1$ and $3$) and we want to see ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
A simple algebra question involving a sum of a series I'm sorry if the question is silly, but I wondered if the next equation is right:
$$\sum\limits_{k = 1}^n {2^{n-k}}k = \sum\limits_{k = 1}^n k \sum\limits_{k = 1}^n {2^k}$$
And if not, is there a way to represent it without the sum $\sum\limits_{k = 1}^n {2^{n-k}}k$... | At first note that left-hand side and right-hand side of the equation are different.
The LHS is
\begin{align*}
\sum_{k=1}^nk\sum_{k=1}^n2^k&=\left(\sum_{k=1}^nk\right)\left(\sum_{k=1}^n2^k\right)\\
&=(1+2+3+\cdots+n)(2^1+2^2+2^3+\cdots+2^n)
\end{align*}
whereas the RHS is
\begin{align*}
\sum_{k=1}^n 2^{n-k}k=1\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate the sum of series $\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$ Evaluate the sum of series
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$$
I have tried two methods:
1) using power series
2) using partial sums
but I can't find the sum.
1) Using power series:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{... |
The way to proceed is to exploit the telescoping series as illustrated by @MarkusScheuer. But, ...
If one really wants to use the method proposed in the OP, then writing
$$\frac{1}{k(4k^2-1)}=\int_0^1 x^{k(4k^2-1)-1}\,dx$$
won't lead to tractable way forward since $\sum_{k=1}^K x^{k(4k^2-1)-1}$ is not a geometric ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
How can I derive the conditions of Positive semidefinite cone in $2\times2$ matrix. By the definition, in order for $X$ to be positive semidefinite cone in $S^2$, it should satisfy that
\begin{equation}
X=\left[
\begin{array}{cc}
x & y \\
y & z
\end{array}
\right]\in S_+^2 \quad\Longleftrightarrow\quad x\ge0,\quad z\ge... | This is essentially the implication from the Sylvester's criterion extended to the case of positive semi-definite matrices.
All of the leading principal minors must be nonnegative. So you immediately get $x\ge 0, x z\ge y^2, z\ge 0$.
Following your proof for $2\times2$ matrices, take consequently $\alpha = 0$ and $\bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How do I solve the differential equation: $ \frac{dy}{dx} = \frac{x+y}{x-y} $? $$ \frac{dy}{dx} = \frac{x+y}{x-y} $$
I have tried this problem so long... such as
please help..
| This is a dimensionally homogenous ODE.
Let $u = \frac{y}{x}$
Then we have $\frac{du}{dx} = \frac{x\frac{dy}{dx} - y}{x^{2}}$
So $\frac{du}{dx} = \frac{\frac{dy}{dx} - u}{x}$
$\frac{dy}{dx} = x\frac{du}{dx} + u$
Substituting this in:
$x\frac{du}{dx} + u = \frac{x + ux}{x-ux}$
$x\frac{du}{dx} + u = \frac{1+u}{1-u}$
$x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluate: $\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$ Here is what I did:
$\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$
Put $x^2=\tan (\theta )$
Then: $x=\sqrt{\tan (\theta )}$ and $dx=\frac{1}{2} \tan ^{-\frac{1}{2}}(\theta ) \sec ^2(\theta )d\theta$
So the integral ... | $\displaystyle J=\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$
Perform the change of variable $y=\dfrac{1-x^4}{1+x^4}$,
$\displaystyle J=\dfrac{1}{2^{\tfrac{9}{4}}}\int_0^1 \dfrac{x^{\tfrac{3}{4}}}{(1-x)^{\tfrac{3}{4}}}dx=\boxed{\dfrac{1}{2^{\tfrac{9}{4}}}\text{B}\left(\dfrac{1}{4},\dfrac{7}{4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Did I do it correctly? What's the image and the basis of the image of this linear mapping?
$A$ is the linear mapping $f(x)= Ax,\mathbb{R} \rightarrow \mathbb{R}$
$$f\left( \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}
\end{pmatrix}\right)= \begin{pmatrix} x_{2}-x_{3}\\
x_{1}+3x_{2}-2x_{3}\\ x_{1}-4x_{2}+5x_{3} \end{pmatri... | You start with the matrix
$$
f =
\left[
\begin{array}{rrr}
0 & 1 & -1 \\
1 & 3 & -2 \\
1 & -4 & 5 \\
\end{array}
\right]=
\left[
\begin{array}{rrr}
f_{1} & f_{2} & f_{3}
\end{array}
\right].
$$
This looks like an academic exercise, so we look for trivial combinations of rows or columns. We see the column property... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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... | You seek the probability that $~2b < 10+a$ when selecting the values with replacement and no bias.
Let us see: You have listed values of $b$ that satisfy this for every $a$, counted them, and compared as a ratio of the size of the sample space.
Yes, that is okay; you have arrived at the correct answer by a valid proces... | {
"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": 0
} |
Congruence Modulo Powers of Primes I need help with the following:
Let $p$ be an odd prime and $x, y \in \mathbb{Z}$. Then $x \equiv y$ mod $p^k \implies x^p \equiv y^p$ mod $p^{k+1}$.
I know that I can write $x^p - y^p$ as $(x - y)F(x,y)$ for some ugly polynomial $F$, but I'm not sure whether that's too much help.
Ta
| Suppose $x \equiv y \pmod {p^k}$, for some positive integer $k$.
Note the identity
\begin{align*}
x^p - y^p &= (x - y)
\left(
x^{p-1} + x^{p-2}y + \cdots + xy^{p-2} + y^{p-1}
\right)\\[4pt]
&=(x - y)
\left(
\sum_{i=0}^{p-1}x^{p-1-i}y^i
\right)
\\[4pt]
\end{align*}
Then
\begin{align*}
&x \equiv y \pmod{p^k}\\[4pt]
\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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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$... | Under the assumption that the integer ratio is positive:
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
LEMMA
Given integers $$ M \geq m > 0, $$ along with positive integers $x,y$ with
$$ x^2 - Mxy + y^2 = m. $$
Then $m$ is a square.
PROOF.
First note that we cannot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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Polynomial form of $\det(A+xB)$
Let $A$ and $B$ be two $2 \times 2$ matrices with integer entries. Prove that $\det(A+xB)$ is an integer polynomial of the form $$P(x) = \det(A+xB) = \det(B)x^2+mx+\det(A).$$
I tried expanding the determinant of $\det(A+xB)$ for two arbitrary matrices, but it got computational. Is the... | $$\begin{pmatrix}a_{11}&&a_{12}\\a_{21}&&a_{22}\end{pmatrix}+x\begin{pmatrix}b_{11}&&b_{12}\\b_{21}&&b_{22}\end{pmatrix}=\begin{pmatrix}a_{11}+xb_{11}
&&a_{12}+xb_{12}\\a_{21}+xb_{21}&&a_{22}+xb_{22}\end{pmatrix}=C$$
$$\det(C)=
a_{11}a_{22}+a_{11}xb_{22}+a_{22}xb_{11}+x^2b_{11}b_{22}-
a_{21}a_{12}-a_{21}xb_{12}-a_{12}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2189714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Uniform continuity : stuck at an intermediate step $$f(x)= \frac{x}{1+x^2}, \, x \in \mathbb{R}$$
I have proved till the following step. I can't understand how do I proceed further. Please help $$f(x)\le|(x-y)||1-xy|$$
| For all $x \in \mathbb{R},$ we have
$$f'(x) = \frac{1 - x^2}{(1+x^2)^2} $$
For $|x| \leqslant 1$ we have
$$|f'(x)| = \frac{1 - x^2}{(1+x^2)^2} \leqslant \frac{1}{(1+x^2)^2} \leqslant 1,$$
and for $|x| > 1$ we have $|f'(x)| \to 0$ as $x \to \pm \infty$ and local maxima at $x = \pm \sqrt{3}$ where $f'(\pm \sqrt{3}) = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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... | Since all normal lines intersect at a common point you have
$$
-(b+a)2ab=-(c+a)2ac=-(c+b)2bc,
$$
which means
$$
b^2a+a^2b=c^2a+a^2c=c^2b+b^2c.
$$
Now if $a=0$
$$
c^2b+b^2c=0\Rightarrow c=-b.
$$
If $a\neq0$,since $b\neq c$, from the first equality
$$
(b^2-c^2)a=-a^2(b-c) \Rightarrow b+c=-a.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is the probability that if $3$ of $12$ lockers are selected at random that at least two of the selections are consecutive? Three persons randomly choose a locker among 12 consecutive lockers.
What is the probability that at least two of the lockers are consecutive?
This is what I came up with:
$|S| = C(12,3)$ all... | Your answer is incorrect. Notice that
$$\frac{\binom{11}{2}\binom{10}{1}}{\binom{12}{3}} = \frac{55 \cdot 10}{220} = \frac{550}{220} = \frac{5}{2} > 1$$
Method 1: There are $12 \cdot 11 \cdot 10 = 1320$ possible ways for three people to choose their lockers.
Now, let's count arrangements in which at least two peop... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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LU-decomposition of A I have:
$A=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
4 & -2 & 7 & 7 & 6
\\
2 & -1 & 20 & 9 & -8
\end{bmatrix}$
and I'm asked to LU-decomposition A, then solve $Ax=0$.
What I did:
With these steps:
$1)$ $R2=R2-2\times R1$
$2)$ $R3=R3-R1$
I got:
$U=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
0 & 0 & 3 & 1 & -... | $A=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
4 & -2 & 7 & 7 & 6
\\
2 & -1 & 20 & 9 & -8
\end{bmatrix}$
With these steps:
$1)$ $R2=R2-2\times R1$
$2)$ $R3=R3-R1$
$3)$ $R3=R3-6\times R2$
I got:
$U=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
0 & 0 & 3 & 1 & -2
\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}$
$L=\begin{bmatrix}
1 & 0 & 0
\\
2 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find closed form of Wallis's product type $\prod_{n=1}^{\infty}\left(\prod_{k=0}^{m}(n+k)^{{m\choose k}(-1)^k}\right)^{(-1)^n}$ Wallis's product
$${2\over 1}\cdot{2\over 3}\cdot{4\over 3}\cdot{4\over 5}\cdots={\pi\over 2}\tag1$$
Generalised of Wallis's product type
$$\prod_{n=1}^{\infty}\left(\prod_{k=0}^{m}(n+k)^{{... | Using $~\displaystyle \prod\limits_{n=1}^M (x+n) \approx \frac{M!M^x}{\Gamma(x+1)}~$ for large $M$ we get with $~m\geq 1~$ :
$$\prod_{n=1}^{\infty}\left(\prod_{k=0}^{m}(n+k)^{{m\choose k}(-1)^k}\right)^{(-1)^n} = 2^{\delta_{m,1}} \prod\limits_{k=1}^m {\binom {k}{k/2}}^{(-1)^k {\binom m k}}$$
where $~\delta_{i,j}~$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the last two digits of $47^{89}$ Find the last two digits of the number $47^{89}$
I applied the concept of cyclicity
$47\cdot 47^{88}$
I basically divided the power by 4 and then calculated $7^4=2401$ and multplied it with 47 which gave me the answer $47$ but the actual answer is $67$. How?
| \begin{align}
47^2 &= 40^2 + 2\times 40\times 7 + 7^2 \equiv 9 \pmod{100},\\
47^4 &\equiv 9^2 \equiv 81 \pmod{100}, \\
47^8 &\equiv 81^2 \equiv 61 \pmod{100}, \\
47^{16} &\equiv 61^2 \equiv 21 \pmod{100}. \\
\end{align}
At this point, if we recognize how convenient the results
$47^4 \equiv 81 \pmod{100}$ and $47^{16} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How can I prove this trigonometric equation with squares of sines? Here is the equation:
$$\sin^2(a+b)+\sin^2(a-b)=1-\cos(2a)\cos(2b)$$
Following from comment help,
$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$
$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \c... | On RHS we have,
$$
1-\cos(2a)\cos(2b) = 1-\cos^2a\cos^2b+\cos^2a\sin^2b+\sin^2a\cos^2b-\sin^2a\sin^2b
= (\cos^2a\sin^2b+\sin^2a\cos^2b)+1-\cos^2a+\sin^b\cos^2a-\sin^2a\sin^2b
= (\cos^2a\sin^2b+\sin^2a\cos^2b)+\sin^2b\cos^2a+\sin^2a-\sin^2a\sin^2b
= 2(\cos^2a\sin^2b+\sin^2a\cos^2b)
= LHS
$$
Hence proved
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 1
} |
Tan function and isosceles triangles I have a non-right-angled isosceles triangle with two longer sides, X, and a short base Y.
I know the length of the long sides, X.
I also know the acute, vertex angle opposite the base Y, let's call it angle 'a'
I have been told I can calculate the length of the base Y by:
Y = tan(a... | bisect angle $a$
$\sin \frac a2 = \frac {y}{2x}\\
\cos \frac a2 = \frac {\sqrt{4x^2-y^2}}{2x}\\
\tan \frac a2 = \frac {y}{\sqrt{4x^2-y^2}}\\
\tan a = \frac {2\tan\frac a2}{1-\tan^2 \frac a2}\\
\tan a = \frac {2y}{\sqrt{4x^2-y^2}(1-\frac {y^2}{4x^2-y^2})}\\
\tan a = \frac {y\sqrt{4x^2-y^2}}{2x^2-y^2}$
$\frac yx$ is smal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solve $f(x) = f(\lfloor x/2 \rfloor) + f(\lfloor x/3 \rfloor)$
Solve the following recurrence relation
$$f(1)=1, f(2)=2\\
f(x) = f\left(\left \lfloor \frac{x}{2} \right \rfloor\right) + f\left(\left\lfloor \frac{x}{3} \right\rfloor\right), \forall x \in \mathbb{N}, x \geq 3 $$
I tried the simple ways to solve... | Is $x$ constrained in some particular set?
Note that:
$$
\begin{aligned}
f(2) &= f\left(\left\lfloor \frac{2}{2} \right\rfloor\right) + f\left(\left\lfloor \frac{2}{3} \right\rfloor\right)\\
&=f(1) + f(0)
\end{aligned}
$$
Therefore, $f(0) = f(2) - f(1) = 1$. At the same time, for $x=1$,
$$
\begin{aligned}
f(1) &= f\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Convergence, absolute convergence of an integral $\int_{0}^{\infty}\frac{(x^{4}+2017)*(sin x)^{3}}{x^{5}+2018*x^{3}+1}dx$
How do I prove that it converges and that it does not converge absolutely?
| Recall that $\sin^3(x)=\frac34\sin(x)-\frac14\sin(3x)$. Furthermore, $\left|\int_0^L \sin(x)\,dx\right|\le 2$ for all $L$.
Then, note that for sufficiently large $x$, $\frac{x^4+2107}{x^5+2018x^3+1}$ monotonically decreases to $0$.
Abel's Test (Dirichlet's Test) for improper integrals guarantees convergence of the i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $\displaystyle \prod_{j=1}^{2n}\left(j^j(2n+j)^{j-1}\right)$ is a perfect square
For any positive integer $n$, prove that $\displaystyle \prod_{j=1}^{2n}\left(j^j(2n+j)^{j-1}\right)$ is a perfect square.
Let $f(n) = \prod_{j=1}^{2n}\left(j^j(2n+j)^{j-1}\right)$. Then $$f(1) = 2^4, \quad f(2) = 2^{20} \cdot... | Extracting even powers, we see that, up to a perfect square, this expression is equal to
$$
\prod_{\text{odd }j\le 2n} j \cdot \prod_{\text{even }j\le 2n}(2n+j) = (2n-1)!! \cdot \prod_{k=1}^n(2n+2k) = (2n-1)!! \cdot 2^n \frac{(2n)!}{n!} \\
= (2n-1)!! \cdot 2^{2n} \frac{(2n)!}{(2n)!!} = \bigl(2^n (2n-1)!!\bigr)^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ I know there are various methods showing that $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, but I want to know how to derive it from letting $t\rightarrow 0^{+}$ for the following identity:
$$\sum_{n=-\infty}^{\infty}\frac{1}{t^2+n^2}=\frac{\pi}{t}\frac{1+e^... | Note we can write:
$$\frac{\pi}{t}-\frac{1}{t^2}\frac{(-2\pi t)e^{-2\pi t}}{1-e^{-2\pi t}}=\frac{\pi}{t}+\frac{1}{t}\frac{2\pi e^{-2\pi t}}{1-e^{-2\pi t}}=\frac{\pi}{t}\left(1+\frac{2e^{-2\pi t}}{1-e^{-2\pi t}}\right)=\frac{\pi}{t}\frac{1-e^{-2\pi t}+2e^{-2\pi t}}{1-e^{-2\pi t}}\\=\frac{\pi}{t}\frac{1+e^{-2\pi t}}{1-e^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is wrong with this way of solving trig equations? Let's suppose I have to find the values of $\theta$ and $\alpha$ that satisfy these equations:
*
*$\cos^3 \theta$ = $\cos \theta $
*$3\tan^3 \alpha = \tan \alpha$
on the interval $[0; 2 \pi]$.
If I try to solve, for instance, the first equation like this:
$$\co... | Clearly, when $\cos \theta = 0$, then $\cos^3 \theta = \cos \theta$.
So, in the interval $[0, 2\pi]$, we get $\theta \in \left\{\frac 12 \pi, \frac 32 \pi \right\}$
If $\cos \theta \ne 0$, then we can divide both sides by $\cos \theta$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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The integral $\int_{-\infty}^\infty \frac{e^{-x^2}}{(1+e^x)^2} dx$. Let
$$T(n) = \int_{-\infty}^\infty \frac{e^{-x^2}}{(1+e^{x})^n} dx.$$
We have that
$$ T(0) = \sqrt{\pi} \text{ and } T(1) = \frac{\sqrt{\pi}}{2}$$
and also that
$$ T(3) = \tfrac{3}{2} T(2) - \frac{\sqrt{\pi}}{4}.$$
Can we find a closed form for $T(2)$?... | An expression for $T(2)$ as a series of Bernoulli numbers:
\begin{align}
T(2) &= \int_{-\infty}^\infty \frac{e^{-x^2}}{(1+e^{x})^2} d\,dx\\
&=\frac{1}{4}\int_{-\infty}^\infty \frac{e^{-x^2-x}}{\cosh^2 \frac{x}{2}} \,dx\\
&=\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2-2x}}{\cosh^2 x} \,dx\\
&=\frac{1}{2}\int_{-\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Proving $\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$.
Prove the following identity:
$$\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$$
How can I express $\cos(4\theta) $ in other terms?
| $$\cos(4\theta)=2\cos^2(2\theta)-1\to\\ \cos(4\theta) - 4\cos(2\theta)=2\cos^2(2\theta)- 4\cos(2\theta)-1\to\\
\cos(4\theta) - 4\cos(2\theta)=2(1-2\sin^2(\theta))^2-4(1-2\sin^2(\theta))-1=8\sin^4(\theta)-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2218085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Need direction for solving integral
Evaluate: $$\int \frac{\sin^2x}{\cos^2x +4}dx.$$
I tried to this things:
First
$$\tan\left(\frac{x}{2}\right) = t$$
$$dx = \frac{2~dt}{1+t^2}$$
$$\sin x= \frac{2t}{1+t^2}$$
$$\cos x= \frac{1-t^2}{1+t^2}$$
Tried also but this is same thing?
$$\tan\left(\frac{x}{2}\right) = t$$
There... | substituting $t=\tan x$, with $dx=\frac{dt}{1+t^2}$ gives $$\int\frac{t^2}{(5+4t^2)(1+t^2)}dt=\int\frac{5}{5+4t^2}-\frac{1}{1+t^2}dt$$
So you end up with $$\frac{\sqrt{5}}{2}\arctan\left(\frac{2}{\sqrt{5}}\tan x\right)-x+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How to know multipliers in solving PDE Equations? Solve $$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dz}{x+y}.$$ Solve the similtaneous equation by using method of multipliers. How can we choose these multipliers? Is there any specific method to know these multipliers?
| Try to find triplets (multipliers) $l,m,n$ such that in,
$$\frac{dx}{y+z}=\frac{dy}{z+x}=\dfrac{dz}{x+y}=\dfrac{l dx+m dy+n dz}{l(y+z)+m(z+x)+n (x+y)}$$
make the denominator vanish
Set $l=y-z,m=z-x,n=x-y$
Checking for the denominator to vanish: $(y-z)(y+z)+(z-x)(z+x)+(x-y)(x+y)=y^2-z^2+z^2-x^2+x^2-y^2=0$
$$\frac{dx}{y+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the minimum value of $|z+\frac{1}{2}|$ for $|z|\geq 2$ Find the minimum value of $|z+\frac{1}{2}|$ for $|z|\geq 2$.
Is the following approaches correct ?
$\left|z+\frac{1}{2} \right|=\sqrt{(x+\frac{1}{2})^2+y^2}=\sqrt{x^2+y^2+x+\frac{1}{4}}$. What will be the next?
Update:
$|z|\geq 2 \implies x^2+y^2... | Use the triangle inequality:
$$||a|-|b|| \leq |a+b| \leq |a|+|b|.$$
You need only the first part with $a=z$ and $b=1/2$. So
$$ |z+\frac12| \geq ||z|-\frac12| = |z| - \frac12 \geq 2 - 0.5 = 1.5$$
It is enough to notice, that for $z=-2$ we have $|z|=2$ and $|z+\frac12| = 1.5$.
More about triangle inequality:
https://en.w... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^{2\pi}\frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\ A,B <<1$ I need to evaluate the definite integral $$\int_0^{2\pi}\frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for various}\ A,B \text{; with}\ A,B<<1.$$
Wolfram Alpha provides the followi... | When one of $A$ and $B$ is non-zero, then one way to figure this out for yourself is to put
$$A \colon= r \cos \beta \ \mbox{ and } \ B \colon= r \sin \beta,$$
where
$$r = \sqrt{A^2 + B^2} \ \mbox{ and } \ B \tan \beta = A.$$
Then
$$
\begin{align}
\int_0^{2\pi} \frac{1}{1+ A \sin \theta + B \cos \theta } \mathrm{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Absolute value distance Let $a$ and $b$ be in the interval $[0, 10]$, and let $a_0$ have a distance inferior to $10^{-3}$ from $a$ and $b_0$ a distance inferior to $10^{-3}$ from $b$. How close is $a_0 b_0$ from $ab$?
This has got me stuck. How can I verify this through inequalities?
| Partial answer: Upper bound for $a_0b_0-ab.$
For $a\leq 10-10^{-3}$ and $b\leq 10-10^{-3}$ we have $$a_0b_0<(a+10^{-3})(b+10^{-3})=ab+10^{-3}(a+b)+10^{-6}\leq$$ $$\leq ab+10^{-3}(20-2\cdot 10^{-3})+10^{-6}=$$ $$=ab+10^{-3}(20-10^{-3}).$$ Observe that in this case $a_0b_0-ab$ can be arbitrarily close to $10^{-3}(20-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Cubic equations and relationship between their roots If there exists a cubic equation $x^3 + 2x^2 +3x + 1 = 0$ has the roots $a,b,c$
And it is given that
*
*$\frac{1}{a^3} + \frac{1}{b^3} - \frac{1}{c^3}$
*$\frac{1}{a^3} + \frac{1}{c^3} - \frac{1}{b^3}$
*$\frac{1}{c^3} + \frac{1}{b^3} - \frac{1}{a^3}$
Are the... | I have to sleep, so this is all I have right now. Perhaps I'll update tomorrow? It's too long for a comment anyways.
So, if we assume it's monic, obviously $p$ will be 1. Now, by Vieta's formulas, $-s$ will be the sum of the roots. And the sum of the roots turns out to be:
$\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231662",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the value of $\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$ Find the value of $\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$
My Attempt:
$$\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$$
$$\sin \dfrac {180}{10} - \sin \dfrac {3\times 180}{10}$$
$$\sin 18^\circ - \sin 54^\circ$$
Now,
Let $A=18^\circ$.
$$5A=90^... | $$\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}=\cos72^{\circ}+\cos144^{\circ}=$$
$$=\frac{2\sin36^{\circ}\cos72^{\circ}+2\sin36^{\circ}\cos144^{\circ}}{2\sin36^{\circ}}=$$
$$=\frac{\sin108^{\circ}-\sin36^{\circ}+\sin180^{\circ}-\sin108^{\circ}}{2\sin36^{\circ}}=-\frac{1}{2}.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $$\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge \frac{9}{4}$$
$$\Leftrightarrow \sum \frac{3(a+b)^2+(a-b)^2}{(a-b)^2} ... | $$a^2-2ab+b^2\le a^2+ab+b^2$$
$$1 \le \frac{a^2+ab+b^2}{(a-b)^2}$$$$1\le\frac{a^3-b^3}{(a-b)^3}$$$$\frac{9}{4}<3\le\sum_{cyc}\frac{a^3-b^3}{(a-b)^3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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'... | Hint....From what you have obtained, $x^2+3xy=3$ make $y$ the subject and substitute this into the equation $2x^2+y^2=6$ and you will obtain a quadratic in $x^2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the rational canonical form of a matrix from its minimal and characteristic polynomials
What is the rational canonical form of $A$?
$$A=\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
2 & 3 & -1 & 4\\
1 & 1 & -1 & 3\\
\end{bmatrix}$$
I found that the minimal polynomial $m_... | If the first matrix is the rational form of $A$, we should have $\dim \ker (A - I) = 3$ (because this is true for the rational form and so it should be true for $A$ as well) while if the second matrix is the rational form of $A$, we should have $\dim \ker(A - I) = 2$. Just check which of those two options holds for $A$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\int x \sqrt{1- x^2 \over 1 + x^2}dx$.
$$J =\int x \sqrt{1- x^2 \over 1 + x^2}dx$$
Substituting $u = \sqrt{1 + x^2}$
$$J = \int \sqrt{2 - u^2} du = \sqrt{2}\int \sqrt{1 - \left({u \over \sqrt{2}}\right)^2} du$$
Now substituting $\sin z = u/\sqrt{2}$
$$J = 2\int \cos^2 z dz = z + (\sin 2z)/2 + C = \arcsin\l... | Note that $$\frac{d}{dx}\arcsin(x^2)=\frac{2x}{\sqrt{1-x^4}}$$ and $$\frac{d}{dx}2\arcsin\left(\frac{\sqrt{1+x^2}}{\sqrt 2}\right)=\frac{2x}{\sqrt{1-x^4}}$$
So, the error in the OP is in the term $\arcsin\left(\frac{\sqrt{1+x^2}}{2}\right)$, which should be instead $\arcsin\left(\frac{\sqrt{1+x^2}}{\sqrt 2}\right)$ and... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Decide if the series $\sum\frac{4^{n+1}}{3^{n}-2}$ converges or diverges and, if it converges, find its sum
Decide if the series $$\sum_{n=1}^\infty\frac{4^{n+1}}{3^{n}-2}$$ converges or diverges and, if it converges, find its sum.
Is this how you would show divergence attempt:
For $n \in [1,\infty), a_n = \frac{4^{n... | For $n\geq 1$ we have $3^n>3^n-2>0$ so $0<3^{-n}<(3^n-2)^{-1}$ so $4^{n+1}(3^n-2)^{-1}>4^{n+1}3^{-n}=4 (4/3)^n>4.$ If the terms of the series do not converge to $0$ then the sum cannot be convergent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How do I compute the taylor series for $(1+x+x^2)^\frac{1}{x}$ I tried rewriting $(1+x+x^2)^\frac{1}{x}$ as $e^{\frac{1}{x}\ln(1+x+x^2)}$ and then computing the taylor series of $\frac{1}{x}$ and $\ln(1+x+x^2)$ but I'm still not getting the correct answer..
| This appears to be the approach you used, but without seeing your work, it is hard to tell where you were having trouble.
Use the series $\log(1+x)=x-\frac{x^2}2+\frac{x^3}3+\cdots$
$$
\begin{align}
\frac1x\log\left(1+x+x^2\right)
&=\frac1x\left(x+x^2-\frac{x^2+2x^3+x^4}2+\frac{x^3+3x^4+3x^5+x^6}3+\cdots\right)\\
&=1+\... | {
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"url": "https://math.stackexchange.com/questions/2240756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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What is $x$ in the formula? What is $x$ (assume it's integer) in $512p+ 1 = x^3$, where $p$ is a prime number.
Attempt:
$$a^3 - b^3 = (a-b) (a^2 + ab+b^2) \Longrightarrow 512p = x^3 -1 = (x-1)(x^2+x+1).$$
Here I got stuck. Am I suppose to plug in $p$ and try it one by one?
What about $16p + 1 = x^3$??
| $512p + 1 = x^3$
$512p = x^3 -1$
$512p = (x-1)(x^2 + x + 1)$
$8^3p = (x-1)(x^2 + x + 1)$
If $x$ is even then $x^2 + x + 1$ is odd, and $x-1$ is odd and LHS is odd. That's impossible.
So $x$ is odd. So $x^2 + x + 1$ is odd. So $512|x-1$
$p = \frac {x-1}{512}(x^2 + x + 1)$.
If $x \le 0$ then $ \frac {x-1}{512}(x^2 + x... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\lim_{(x,y)\to (0,0)} (x^4+y^4)/(x^2+y^2)=0$ by definition I need to prove by definition (and nothing else) that
$$\lim_{(x,y) \to (0,0)}\frac{x^4+y^4}{x^2+y^2} = 0.$$
I've been stuck on this for almost an hour with no luck, and ran out of ideas.
Can anyone help or give a hint?
| These things are always nicer in polar form...
$$\frac{x^4+y^4}{x^2+y^2} = \frac{r^4 (\sin^4 \theta + \cos^4 \theta)} {r^2} = r^2 (\sin^4 \theta + \cos^4 \theta) \leq 2r^2$$
Let's assume $\|(x,y)\| < \delta$, that is $(x,y)$ is inside the open disk around $0$ with a radius of $\delta$.
Then using the above inequalit... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y\rfloor+\lfloor x+y\rfloor$
Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.
If anybody could post a simple solution (no complicated abstract ... | HINT
Let $x = n + r$ with $n$ an integer and $0 \le r <1$
Similarly, $y = m + s$ with $m$ an integer and $0 \le s < 1$
Now there are 4 cases to consider:
*
*$0 \le r < \frac{1}{2}$ and $0 \le s < \frac{1}{2}$
*$0 \le r < \frac{1}{2}$ and $\frac{1}{2} \le s <1$
*$\frac{1}{2} \le r < 1 $ and $0 \le s < \frac{1}{2}$
... | {
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"answer_id": 0
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Show that if $p$ and $q$ are primes $\equiv 3$(mod $4$) then at least one of the equations $px^{2}-qy^{2} = \pm 1$ is soluble in integers $x, y$. So far we have only talked about equations in the form $x^{2} - dy^{2} = \pm 1$ and I'm unsure how to handle the specific coefficients. We have normally found $\sqrt{d}$ and ... | You have a primitive solution to $x^2-pqy^2=1$ ($x$, $y$ positive, $y$ minimal). Then $x$ is odd and $y$ is even (think modulo $4$). Also
$$x^2-1=(x+1)(x-1)=pqy^2$$
so that
$$\frac{x+1}2\frac{x-1}2=pq\left(\frac{y}{2}\right)^2.$$
As $(x\pm 1)/2$ are coprime integers then we have one of the following
*
*$(x+1)/2=u^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"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:
$$P(k): 3^{2k+1} + 2^{k-1},\ P(k+1): 3^{2(k+1)+1} + 2^{(k+1)-1}.$$
$$3^{2k+3} + 2^k = 9 \times 3^{2k+1} + 2^{k-1} \tim... | $$P(k)=3^{2k+1}+2^{k-1}=7 \lambda (let)$$
Now, For $(k+1)$,
$$9*3^{2k+1}+2^{k-1}.2$$
Put $2^{k-1}$ form first equation i wrote
$$7.(3^{2k+1}+2\lambda)$$
Hope your doubt is solved?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Showing $\frac{1}{x^3+x}$ is continuous at $x=1$ I've been attempting to use the definition of continuity and felt a little uncertain about my working on this question.
We define $\epsilon>0$ and choose $\delta $ s.t .....
Let $\lvert x-1\rvert < 1 $:
$\lvert \frac{1}{x^3+x}-\frac{1}{2} \rvert = \lvert \frac{2-x^3-x... | Note that we have
$$\left|\frac{1}{x(x^2+1)}-\frac12\right|=\left|\frac{x^2+x+2}{2x(x^2+1)}\right|\,|x-1|$$
Now, we need to bound $x$ away from $0$. So, let us first restrict $x$ so that $1/2<x<3/2$. Then, we see that
$$\left|\frac{x^2+x+2}{2x(x^2+1)}\right|\le \frac{(3/2)^2+(3/2)+2}{((1/2)^2+1)}=\frac{23}{5}$$
Henc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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why$2 (-1 + 2^{1 + n})$ is the answer to the recurrence relation $a_{n}=2a_{n-1}+2$? $a_{0}=2$
$a_{1}=2(2)+2$
$a_{2}=2(2(2)+2)+2$
$a_{3}=2(2(2(2)+2)+2)+2$
$a_{4}=2(2(2(2(2)+2)+2)+2)+2$
$a_{5}=2(2(2(2(2(2)+2)+2)+2)+2)+2$
To simplifiy
$a_{6}=2^{6}+2^{5}...2^{1}$
so my answer is
$a_{n}=2^{n+1}+2^{n}+...2^{1}$
The correct... | Let $a_n=2^nb_n$. Then
$$
a_{n}=2a_{n-1}+2\\\Downarrow\\2^nb_n=2^nb_{n-1}+2\\\Downarrow\\b_n=b_{n-1}+2^{1-n}
$$
Therefore,
$$
\begin{align}
b_n
&=b_0+\sum_{k=0}^{n-1}2^{-k}\\
&=b_0+\frac{1-2^{-n}}{1-2^{-1}}\\[9pt]
&=b_0+2-2^{1-n}
\end{align}
$$
and
$$
2^{-n}a_n=a_0+2-2^{1-n}\\
\Downarrow\\
\begin{align}
a_n
&=2^na_0+2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Stars and Bars with a maximum number per tuple I have 4-tuples of numbers, each consisting of 1 to 10. How do I find the number of permutations possible to reach a total sum of 21? I tried using the formula given in https://en.wikipedia.org/wiki/Binomial_coefficient and plotted it against a graph, problem is, my graph ... | An approach using generating functions:
In order to count up the number of solutions that sum to $n$, we can look at the coefficient of $x^n$ in the generating function $f(x)$:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=1}^{10}x^n\right)^4$$
The $[x^n]$ operator refers to the coefficient of $x^n$.
Use the identity $\displays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Remainder when dividing power series I know how long division of polynomials works. For example I can use it to divide $x^2+1$ into $x$ to get
$$x=(x^2+1)\frac{1}{x}-\frac{1}{x}$$
with remainder $R=-1/x$, so that
$$\frac{x}{x^2+1}=\frac{1}{x}-\frac{1/x}{x^2+1}.$$
My question is, suppose $f$ and $g$ have infinite power ... | No, the remainder is not $0$, but another power series. For example,
$$
\begin{align}
\frac{\sin(x)}{\cos(x)}
&=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+O\!\left(x^9\right)}{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+O\!\left(x^8\right)}\\[6pt]
&=x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+O\!\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $T$ is a Linear Transformation , if $T$ is defined by $T(A)=XA-AX$ Let $T : M_{2\times 2} \to M_{2\times 2}$ be defined by
$$
T(A) =
\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} A
− A \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}.
$$
How do I prove that $T$ is a linear transformation and then find the basis... | Linearity is easy to prove.
Now let $A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}$ then $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 1 &2\end{pmatrix}=\begin{pmatrix} 2a+b & a+2b \\ 2c+d & c+2d \end{pmatrix}=D_1$
And $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}A=\begin{pmatrix} 2a+c & 2b+d \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Floors Complicated Proof Problem:
Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.
I have my proof down below. Its a little complicated though.
This is not a duplicate of the same question on another page on math.stackexcha... | There is a simple solution to this, and it involves a little casework.
Case 1: The fractional parts of $x$ and $y$ are both less than $0.5$. Then
$$\lfloor 2x\rfloor=2\lfloor x\rfloor$$
$$\lfloor 2y\rfloor=2\lfloor y\rfloor$$
$$\lfloor x+y\rfloor=\lfloor x\rfloor + \lfloor y\rfloor$$
and the two sides of the inequali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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On the differentiability of $\frac{\sin(x^2)+\cos(y^2)}{\sqrt{x^2+y^2}}$ Does this function $f:\mathbb{R^2}\to\mathbb{R}$ defined by:
$$\frac{\sin(x^2)+\cos(y^2)}{\sqrt{x^2+y^2}}$$
differentiable at the origin? Thank you
| The function
$$
f(x) = \frac{\sin \left(x^2\right)+\cos \left(y^2\right)}{\sqrt{x^2+y^2}}
$$
is not differentiable at the origin.
The iterated limit demonstrates the problem.
$$
\begin{align}
\lim_{\color{red}{x}\to 0} f(x,y) &= \frac{\cos \left(y^2\right)}{\sqrt{y^2}} \\[3pt]
\lim_{\color{blue}{y}\to 0} \left( \li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve the Differential equation $x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$ Solve the Differential equation $$x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$$
i reduced the equation as
$$x^3\frac{dy}{dx}=y^3\left(1+\sqrt{1+\left(\frac{x}{y}\right)^2}\right)$$ $\implies$
$$\frac{x^3}{y^3}\frac{dy}{dx}=\left(1+\sqrt{1+\left(\frac... | The solution of the ODE is presented below on parametric form. The explicit form can be obtained but involves the roots of a cubic polynomial equation which would lead to a big formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Measure of a subset of $\mathbb{R}^2$ I'd like to know whether the following integral is right or wrong. I have to calculate the measure of $A$, where
$$
A=\left \{ (x,y): \frac{x^2}{3}+y^2\leqslant 1, \, x\geqslant 0, \, -x\leqslant y\leqslant x \right \}
$$
using polar coordinates $\Phi:\left \{ x= \sqrt{3}\rho \cos ... | I don't believe $B$ is correctly defined. $\theta$ should run between $-\pi/3$ and $\pi/3$. Notice that for $\theta=\pi/3$, $x=\frac{\sqrt{3}}{2}\rho$ and $y=\frac{\sqrt{3}}{2}\rho$.
Here is another approach to verify the answer. You want to intersect two regions
1)the interior of the ellipse $x^2/3+y^2=1$
2)the cone ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Equation of common tangent Equation of common tangent of circle $x^2 + y^2 -8x =0$ and hyperbola with $x^2/9 - y^2/4 =1.$
I tried this but was left with lengthy solutions can u please explain with easy solutions.
| Sketch (this answer is to find common tangent lines between the two objects). If you just want the intersections, substitute $y^2$ from the second equation into the first equation:
*
*First, observe that by implicit differentiation on the hyperbola's equation, we get
$$
\frac{2}{9}x-\frac{1}{2}y\frac{dy}{dx}=0.
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ $A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ and $\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$. Then $A+2B$ is equal to:
$1$. $\dfrac {\pi}{4}$
$2$. $\dfrac {\pi}{3}$
$3$. $... | The first condition gives $3\cos2A+2\cos2B=3$.
The second condition gives $3\sin2A=2\sin2B$ or
$$9\sin^22A=4\sin^22B$$ or
$$9(1-\cos^22A)=4(1-\cos^22B)$$ or
$$9\cos^22A-4\cos^22B=5$$ or
$$(3\cos2A+2\cos2B)(3\cos2A-2\cos2B)=5$$ or
$$3(3\cos2A-2\cos2B)=5$$ or
$$3\cos2A-2\cos2B=\frac{5}{3},$$
which after summing with $$3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Exercise 5.3 in Calculus Made Easy: are these answers equivalent? Exercise 5.3 in Calculus Made Easy, by Silvanus Thompson, is to find $\mathrm d\over \mathrm dx$ when the following relationship holds ($a$ and $b$ are both constants):
$$ay + bx = by - ax + (x + y)\sqrt{a^2 - b^2}$$
I tried differentiating both sides of... | $${\sqrt{a^2 - b^2} - a - b \over a - b - \sqrt{a^2 - b^2}}=\frac{\sqrt{a+b}\left(\sqrt{a-b}-\sqrt{a+b}\right)}{\sqrt{a-b}\left(\sqrt{a-b}-\sqrt{a+b}\right)}=\frac{\sqrt{a+b}}{\sqrt{a-b}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
calculate polynomial equation from roots I want to derive a polynomial equation from the following roots:
$$\left\{-1,1+\sqrt5,1-\sqrt5\right\}$$
Below is my attempt.
\begin{align}f(x)&=(x+1)( (x-1) - \sqrt5 ) ( (x-1) + \sqrt5 )\\
&=(x+1)( (x-1)^2 - 5)\\
&=(x+1)(x^2 - 2x - 4)\end{align}
And the answer in the book is $... | When we have the roots $\{a,b,c\}$, we can write $$f(x)=(x-a)(x-b)(x-c)$$We have the roots $\{-1,1+\sqrt 5,1−\sqrt5\}$ so we can write the equation in the following way:
\begin{align}f(x)&=(x-(-1))(x-(1+\sqrt5))(x-(1-\sqrt5))\\
&=(x+1)(x-1-\sqrt5)(x-1+\sqrt5)\\
&=(x+1)(x^2-x+x\sqrt5-x+1-\sqrt5-x\sqrt5+\sqrt5-5)\\
&=(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove this property involving the floor function? How does one prove that $$\lfloor x \rfloor = \lfloor x/2 \rfloor + \lfloor (x+1)/2 \rfloor $$
where $\lfloor x \rfloor$ represents the greatest integer less than or equal to $x$?
I tried to do this by using inequalities but end up with $\lfloor x \rfloor + 1/2$ ... | Let $x = a + b$ where $a \in \mathbb{N}$ and $0 \leqslant b < 1$
We need to prove that
$$\lfloor a + b \rfloor = \left\lfloor \frac{a+b}{2} \right\rfloor + \left\lfloor \frac{a+b +1}{2} \right\rfloor $$
The L.H.S. will be just $a$
Now the R.H.S.
$$\left\lfloor \frac{a+b}{2} \right\rfloor + \left\lfloor \frac{a+b +1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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When is $n\cdot m\cdot [(n-8)\cdot m+4]-4$ an integer square? For which odd integers $n$ and $m$, with $n \ge m$ is $$n\cdot m\cdot [(m-8)\cdot n+4]-4$$ a square?
Part of my efforts so far:
The product of the three factors above is a sum of two squares (the square we wish to find plus $2^2$), thus all three factors are... | For every $m$ such that $-1$ is quadratic residue modulo $m$ and $m>5$ you will have infinitely many $n$ which satisfy that $n m ((m-8) n+4)-4$ is perfect square.
For Example: for $m=13$ we get that $n=\{29,125,1956797,8346845,130248348509,555582723389,8669590571525885,3698069722589
8397,577065287491657635485,24615... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the value of $A$ for the inequality if $x,y,z$ are positive reals and
$$S=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)} \le A$$ find $A$
i have taken L.C.M getting
$$S=\frac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)}$$ dividing by $xyz$ we get
$$S=\frac{2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}{xyz\le... | Without loss of generality, we can consider $x\le y \le z$ or $x \le ax \le abx$ with $a,b\ge1$.
The given inequality becomes equivalent to:
$$S=\frac{x}{x(a+1)\cdot x(ab+1)}+\frac{ax}{x(b+1) \cdot ax(b+1)}+\frac{abx}{x(ab+1) \cdot ax(b+1)} \le A$$
If $x\to0^+$, the LHS $\to+\infty=A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding $\gcd(x^3, x^3+x+1)$ Suppose $f(x) = x^3$ and $g(x) = x^3+x+1$. I am trying to show that the $\gcd(f,g)= 1$ but am running into some trouble...
My attempt: (Division algorithm)
$x^3+x+1 = (1)(x^3)+(x+1)$
$x^3 = x^2(x+1)-x^2$
But I cannot go further than this, what is going wrong here?
| Just keep going. You found that
$$
x^3+x+1 = (1)(x^3) + (x+1)
$$
Now observe that
$$
x^3 = (x^2-x+1)(x+1) - (1)
$$
What can you conclude?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that the geometric images of any complex number defined by $cis \theta$ belong to a circumference of radius 1
Show that the geometric images of any complex number defined by $cis
\theta$ belong to the circumference centered at the origin with radius
1.
My book states that this is proven because $|cis \theta|=... | The modulus or norm of any complex number $z = a + bi$, $a, b \in \Bbb R$, is given by
$\vert z \vert = \sqrt{a^2 + b^2}; \tag{1}$
using the identity
$x^2 - y^2 = (x + y)(x - y), \tag{2}$
which holds in any field (actually, in any commutative ring), and taking $x = a$, $y = bi$, and recalling that $i^2 = -1$, we find
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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why $x^2-7y^2 = 11$ cannot have integer solutions? I have a quadratic equation $x^2-7y^2 = 11$, I understand it's not a pell equation, because $7/11$ is not an integer, but how can I prove it has no integer solutions?
| If there were integers $x,y$ such that
$$ x^2 - 7y^2 = 11 $$
we see that it would follow that:
\begin{align*}
x^2 - 7y^2 &\equiv 11 \pmod{4} \\
x^2 + y^2 &\equiv 3 \pmod{4}
\end{align*}
We see that the congruence classes mod 4 are $\{0,1,2,3\}$. Hence, the quadratic residues are $\{0,1\}$. Since $x^2, y^2 \in \{0,1 \} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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To prove $\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n} \le \frac{1}{\sqrt{3n+1}}$ To prove $$P=\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n}\le \frac{1}{\sqrt{3n+1}}$$
i have written $P$ as
$$P=\frac{(2n)!}{2^{2n}(n!)^2}=\frac{(2n)!}{4^{n}(n!)^2}=\frac{\binom{... | We can use simple induction to find the answer. First, at $n=1$, the inequality holds true. Now let us assume the above for n and prove it for $n+1$.
Let us divide the LHS of the inequality for $n+1$ by LHS of the inequality for n, and do the same with the RHS for $n+1$ and $n$.
$(2n+1)/(2n+2)$ must be lesser than or e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$. If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
$$\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 =... | Here is a different approach: Since $1 + \tan^2\theta = \sec^2\theta$, we have
$$\sec^2\theta - \tan^2\theta = 1$$
Factoring yields
$$(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$$
Since we are given that $\sec\theta + \tan\theta = x$, we obtain
$$x(\sec\theta - \tan\theta) = 1$$
Therefore,
$$\sec\theta - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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How to solve this integral using the method of residues? Using the method of residues, verify the following.
$$\int_0^{\pi} \frac{d \theta}{(3+2cos \theta)^2} = \frac{3 \pi \sqrt{5}}{25}$$
I tried doing this but can't get the correct answer, here is my attempt:
first I substituted $cos \theta = \frac{z+ \frac{1}{z}}{2}... | Note
$$\int_0^{\pi} \frac{d \theta}{(3+2\cos \theta)^2} = \frac12\int_0^{2\pi} \frac{d \theta}{(3+2\cos \theta)^2}.$$
Let $z=e^{i\theta}$ and hence one has
\begin{eqnarray}
&&\int_0^{\pi} \frac{d \theta}{(3+2\cos \theta)^2}\\
&=&\frac12\int_0^{2\pi} \frac{d \theta}{(3+2\cos \theta)^2}\\
&=&\frac12\int_{|z|=1}\frac{1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{1}{r} = \frac{1}{a} + \frac{1}{b}$ for a semicircle tangent within a right triangle
I have a right angled triangle with the sides which are not hypotenuses $a$ and $b$. There is also a semi-circle radius $r$ whose diameter lies on the hypotenuse of the right angles triangle, and sides $a$ and $b$ are tang... | We can write the area of $\triangle ABC$ in two ways: $$Area = \frac{1}{2}\cdot ab$$ $$Area = \frac{1}{2}\cdot a\cdot OP+\frac{1}{2}\cdot b\cdot OQ = \frac{1}{2}\cdot (ar+br)$$
Equate these two equations and divide both sides by $\frac{1}{2}\cdot abr$ to get your result:
$$\frac{1}{2}\cdot ab=\frac{1}{2}\cdot (ar+br) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
GCD of polynomials in $\mathbb Z_5$
I have to find the $\gcd(p(x),h(x))$ where $p(x)=1+2x+x^2+3x^3$ and $h(x)=2+x+x^5+x^6$ are in $\mathbb Z_5$.
To find the solution, I divide $h(x)/p(x)$ and I get $q_1(x)=2x^3+3x^2-1$ and $r_1(x)=3x^2+3x+3$
Should I divide $p(x)/r_1(x)$ ?
I think $r_1(x)$ is irreducible but I don't ... | The GCD doesn't change if you multiply $p$ by $2$, so to get
$$
2+4x+2x^2+x^3
\qquad\text{and}\qquad
2+x+x^5+x^6
$$
The first remainder is $3x^2+x$, but we can multiply it by $2$:
$$
2x+x^2
\qquad\text{and}\qquad
2+4x+2x^2+x^3
$$
The remainder is $4x+2$, but we can multiply it by $4$:
$$
x+3
\qquad\text{and}\qquad
2x+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $n^5+n^4+1$ is composite for $n>1.$
Prove that $f(n)=n^5+n^4+1$ is composite for $n>1, n\in\mathbb{N}$.
This problem appeared on a local mathematics competition, however it looks like there is no simple method to solve it. I tried multiplying it by $n+1$ or $n-1$ and then tried factorizing it, but it was t... | If $\omega$ is a (complex) root of $x^2+x+1$, then $\omega^3 - 1 = (\omega-1)(\omega^2+\omega+1) = (\omega-1)\cdot 0 = 0$, and hence $\omega^3 = 1$.
Observe then that $\omega^5 = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2$ and $\omega^4 = \omega^3 \cdot \omega^1 = 1 \cdot \omega = \omega$, so that $\omega^5+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$a,b,c,d$ are real numbers such that....
Suppost that $a,b,c,d$ are real numbers such that $$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$$
Basically,I've no any idea or tactics to tackle this problem. Any methods? Thanks in advance.
EDITED.
The given hint in the book... | What if $b=0?$
Else $abcd\ne0,$
we have $\dfrac ab=-\dfrac dc=k$(say)
$\implies a=bk,d=-ck$
$$1=a^2+b^2=b^2(1+k^2)\implies b^2=?,a^2=(bk)^2=?$$
$$1=c^2+d^2=c^2(1+k^2)\implies c^2=?,d^2=(-ck)^2=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How can we prove the equality of the following determinants? Assume that $A$ is an $n\times n$ real matrix whose entries are all $1$. Then how can we show the following determinant equality for any $x$?
$\det(A-xI)$=\begin{vmatrix}
1 -x & 1 & 1 & \cdots & 1 \\
1 & 1 -x & 1 & \cdots & 1 \\
... | Just add row 2, row 3,... up to row $n$ to row 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving a rational inequality with an unknown using cases So the question is,
For what values of x is $\dfrac{1}{1 + x}> -1 $ ?
Now, one way to do it is write the inequality relative to zero, and then find the behavior of the graph relative to the zeroes of the numerator and denominator:
$\dfrac{1}{1 + x}> -1 \implie... | I think the question has been well-answered, but in case you were looking for a way to multiply through and not have to worry about these cases, try this:
$$\frac{1}{1+x} > -1$$ Multiplying through by the positive quantity $(1+x)^2$ gives, $$ (1+x)^2 \frac{1}{1+x} > -(1+x)^2$$ $$ 1+ x > -(1+x)^2$$ $$(1+x) + (1+x)^2 > 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The distance from the incenter to an acute vertex of a right triangle I'm seeking an alternative proof of this result:
Given $\triangle ABC$ with right angle at $A$. Point $I$ is the intersection of the three angle lines. (That is, $I$ is the incenter of $\triangle ABC$.) Prove that
$$|CI|^2=\frac12\left(\left(\;|B... | Alternative proof: by Stewart's theorem the length $\ell_c$ of the angle bisector through $C$ is given by
$$ \ell_c^2 = \frac{ab}{(a+b)^2}\left[(a+b)^2-c^2\right] $$
and by Van Obel's theorem and the bisector theorem $\frac{CI}{\ell_c}=\frac{a+b}{a+b+c}$. It follows that, in general:
$$ CI^2 = \frac{ab}{(a+b+c)^2}\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of $(1-\cos x)/(x\sin x)$ as $x \to 0$ Can you please help me solve:
$$\lim_{x \rightarrow 0} \frac{1- \cos x}{x \sin x}$$
Every time I try to calculate it I find another solution and before I get used to bad habits, I'd like to see how it can be solved right, so I'll know how to approach trigonometric l... | Just taking terms of the fraction in question leads to:
\begin{align}
\frac{1-\cos(x)}{x \, \sin(x)} &= \frac{1 - \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\right)}{x^2 \, \left(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots \right)} \\
&= \frac{\frac{1}{2!} - \frac{x^2}{4!} + \cdots}{1 - \frac{x^2}{3!} + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 6
} |
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.