Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find an angle not on the circumference of a circle problem. I know this has to be extremely easy but I'm not going to solve this problem.
The task is to find the angle at point $A$.
Thanks!
| Notice that $\angle EOF+\angle BOC+49^\circ+175^\circ=360^\circ$. It follows that $\angle EOF+\angle BOC=136^\circ$.
$\angle OCA=\dfrac{180^\circ-\angle BOC}{2}$ and $\angle OEA=\dfrac{180^\circ-\angle EOF}{2}$ because $\triangle OBC$ and $\triangle OEF$ are isosceles respectively.
Now $\angle OAC+\angle OCA+\angle BOC... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that the sequence $b_n = 1$, $b_{n+1} = \frac{1}{1+b_n}$ converges. I need to prove that the sequence $b_1 = 1$, $b_{n+1} = \frac{1}{1+b_n}$ converges.
If $\lim_{n\to\infty} b_n = L$ then $\lim_{n\to\infty} b_{n+1} = L$, so $\lim_{n\to\infty} b_{n+1} = \frac{1}{1+\lim_{n\to\infty} b_n} \implies L = \frac{1}{1+L} ... | Try showing that the subsequences of even and odd terms are monotonic and bounded. If this is the case, then both subsequences must converge. From there it shouldn't be too hard to show that they converge to the same thing, and therefore the overall sequence must converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Triangle inequality with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$ Given a triangle with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$. Show that
$$r_{a}^{2}+ r_{b}^{2}+ r_{c}^{2}\geq 3\sqrt{3}. S+ \left ( m_{a}- m_{b} \right )^{2}+ \left ( m_{b}- m_{c}... | There is another proof bases on $R, r, s$ relationship as follows:
Let $R, r, s$ be the circumcircle radius, incircle radius, and semi-perimeter of the triangle. Then $a, b, c, r_a, r_b, r_c$ satisfy the following relationship:
$$ R\ge 2r$$
$$ r_a+r_b+r_c=4R+r $$
$$r_ar_b+r_br_c+r_cr_a=s^2 $$
$$m_a^2+m_b^2+m_c^2=\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $ $ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $
Answer:
$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \\ \Rightarrow \lim_{x \to \infty} [\frac{x^2+1-ax^2-ax-bx-b}{x+1}]=0 \\ \Rightarrow \... | Let $f(x)=\dfrac{x^2+1}{x+1}$. If
$$
\lim_{x\to\infty}(f(x)-ax-b)=0
$$
then also
$$
\lim_{x\to\infty}\frac{f(x)-ax-b}{x}=0
$$
Thus we must have
$$
\lim_{x\to\infty}\left(\frac{x^2+1}{x(x+1)}-a\right)=0
$$
and therefore $a=1$. Now
$$
\frac{x^2+1}{x+1}-x=\frac{x^2+1-x^2-x}{x+1}=\frac{-x}{x+1}
$$
so
$$
\lim_{x\to\infty}(f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Last three digits of $6^{2002}$ Find the last three digits of $6^{2002}$. I did some work and figured out that the last two digits is 36. Can anyone help me with the hundredth digit? By the way, I used modular arithmetic and the recursion method for the tens digit, but it fell short when I attempted to do the hundreds ... | $3^{400}\equiv 1\mod 1000$ so $3^{2002}\equiv 9\mod 1000$.
$2^{100}\equiv 1\mod 125$ so $2^{2002}\equiv 4\mod 125$.
And $2^{2002}\equiv 0\mod 8$. So by Chinese Remainder theorem $2^{2002}\equiv 504 \mod 1000$.
So $6^{2002}\equiv 9*504\equiv 536\mod 1000$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
How to find a real $d \ne 0$ such that the equation $(2-3i) z^2 -(d-1)z + 4+ 3i = 0$ has a real root? I need to find a real value for d so that the equation $(2-3i)z^2 -(d-1)z + 4+3i= 0$ has real roots.
I've tried using the discriminant formulae and then I got
$$\frac {(d-1) \pm \sqrt{d^2 -2 d-67 +24i}}{4-6i} $$
Here... | Even if you don't have gimusi's insight into considering the conjugate equation, it's possible to "plod" through to a solution.
Let $z = x + yi$. Ultimately, we want to set $y=0$ for the real root(s), but for now, let's do the algebra. $z^2 = x^2 - y^2 + 2xyi$, so the equation becomes:
$$(2-3i)(x^2-y^2+2xyi) - (d-1)(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding unit digit of $f(10)$
If we define $$f(x)=\left\lfloor \frac {x^{2x^4}}{x^{x^2}+3}\right\rfloor$$ and we have to find unit digit of $f(10)$
I had tried approximation, factorization and substitutions like $x^2=u$ but it proved of no use. Moreover the sequential powers are feeling the hell out of me. Can someo... | When $x=10$, $\displaystyle \frac {x^{2x^4}}{x^{x^2}+3}=\frac {10^{20000}}{10^{100}+3}$.
Let $y=10^{100}$. $\displaystyle \frac {10^{20000}}{10^{100}+3}=\frac{y^{200}}{y+3}$.
When $y^{200}$ (as a polynomial) is divided by $y+3$, the remainder is $(-3)^{200}=3^{200}$.
So, $\displaystyle \frac{y^{200}}{y+3}=Q(y)+\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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In any triangle, if $\frac {\cos A+2\cos C}{\cos A+2\cos B}=\frac {\sin B}{\sin C}$, then the triangle is either isosceles or right-angled In any triangle, if $\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$, prove that the triangle is either isosceles or right angled.
My Attempt:
Given:
$$\dfrac {\cos... | You are very close to answer.
*
*cross multiply.
*subtract 1 from both side.
*take b-c common from Numerator.
*It's done!! :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\int\frac{1}{x^3+1}dx$ The problem is, as stated:
$$\int\frac{1}{x^3+1}dx$$
I tried using substitution: $t^3 = x^3 + 1$ but didn't get far with that. I also tried setting: $t = x^3 + 1$, with no luck again.
I tried partial decomposition but I didn't know how to integrate $$\int\frac{1}{x^2-x+1}$$ and I kept ... | $$
\begin{aligned}
I &=\int \frac{4}{4 x^2-4 x+1} d x \\
&=4 \int \frac{d x}{(2 x-1)^2+ (\sqrt 3)^2} \\
&=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$ Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$.
Want to make sure that my proof is correct.
Suppose there are rational solutions.
Note $2x^2 + 3y^2 = 1 \iff 2x^2 + 3y^2 = z^2\ |\ x,y.z \in \mathbb{Z}$
Since taking $(\... | As José Carlos Santos was pointing out in the comments, we have
$$2x^2 \equiv z^2 \mod 3$$
But this does not imply your statement that
$$2x^2 \equiv 1 \mod 3$$
If $z$ were divisible by $3$ and we would have
$$2x^2 \equiv 0 \mod 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
In any $\triangle ABC$, prove that: $\frac {\cos B-\cos C}{\cos A +1}=\frac {c-b}{a}$ In any $\triangle ABC$, prove that: $\dfrac {\cos B-\cos C}{\cos A +1}=\dfrac {c-b}{a}$
My Attempt:
$$\begin{align}
\text{R.H.S.}&=\dfrac {c-b}{a} \\[4pt]
&=\frac {a\cos B+b\cos A-a\cos C-c\cos A}{b\cos C+c\cos B} \\[4pt]
&=\dfrac {a(... | For $\triangle ABC$,
$$\frac{\sin\left(A\right)}{a}=\frac{\sin\left(B\right)}{b}=\frac{\sin\left(C\right)}{c}=\frac{1}{k}$$
therefore,
$$k\sin\left(A\right)=a,\:k \sin\left(B\right)=b,\:k\sin\left(C\right)=c$$
\begin{align*}
RHS= &\frac{k\sin\left(C\right)-k\sin\left(B\right)}{k\sin\left(A\right)}\\&=\frac{\sin\left(C\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
proof verification $\frac{3+2\sqrt{6}}{1-\sqrt{6}}$ is an algebraic integer
Is $$\frac{3+2\sqrt{6}}{1-\sqrt{6}}$$
an algebraic integer?
An algebraic integer means an algebraic number in some algebraic number field $K\supset \Bbb Q$ that is the root of a monic polynomial $f\in \Bbb Z[x]$. Here I guess we are in $\Bb... | $$ \beta =\frac{3+2\sqrt{6}}{1-\sqrt{6}}\implies $$
$$\beta= \frac{(3+2\sqrt{6})(1+\sqrt 6)}{-5}\implies $$
$$-5\beta =15+5\sqrt{6}\implies $$
$$ \beta =-3-\sqrt 6$$
$$ \beta ^2 + 6\beta +3=0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Proving determinant with variables I have a problem that asks:
Prove that det$\begin{pmatrix}
1 && 1 && 1 \\
a && b && c \\
a^2 && b^2 && c^2 \end{pmatrix} = (c-a)(c-b)(b-a)$
I was thinking of solving it like normal and see if I can end up with $(c-a)(c-b)(b-a)$
What I did:
det$\begin{pmatrix}
b && c \\
b^2 && c^2 \e... | It might be useful to look for helpful column/row operations before expanding.
Here, subtraction of the first column from the second and the third one simplify the determinant considerably:
$$\det\begin{pmatrix}
1 && 1 && 1 \\
a && b && c \\
a^2 && b^2 && c^2 \end{pmatrix} =
\det \begin{pmatrix}
1 && 0 && 0 \\
a &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Determine the value of $ax^5 + by^5$ given that $a$, $b$, $x$, and $y$ satisfy the following system: Given
$\begin{cases} ax + by = 4 \\ ax^2 + by^2 = 8 \\ ax^3 + by^3 = 17 \\ ax^4 + by^4 = 42 \end{cases} $
Determine the value of $ax^5 + by^5$
Do you have any brilliant idea to solve this problem? By looking at the ri... | You have
$$ (x+y)(ax^2+by^2) = ax^3+by^3+xy(ax+by) \implies 8(x+y) = 17+4xy $$
$$ (x+y)(ax^3+by^3) = ax^4+by^4 + xy(ax^2+by^2) \implies 17(x+y) = 42 + 8xy $$
It follows that
$$ x+y = 8 $$
$$ xy = \frac{47}{4} $$
Also, from the same relations
$$ ax^5 + by^5 = (x+y)(ax^4+by^4) - xy(ax^3+by^3) = 42(x+y)-17xy = \frac{545... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Find Jordan canonical form and basis of a linear operator. Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear operator such that: $T(x,y,z)=(-y-2z,x+3y+z,x+3z)$, I need to find a Jordan canonical form and a basis.
This is what i did:
In the first place, I found the associated matrix to this linear operator in the canon... | Well, $(A - 2 I )^3 = 0,$ but $(A - 2 I )^2 \neq 0.$ So the minimal polynomial and the characteristic polynomial agree, meaning each eigenvalue occurs in just a single block. Indeed
$$
(A - 2 I )^2 =
\left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 0 & 0 \\
-1 & -1 & -1
\end{array}
\right)
$$
Now find some $u$ such that $(A ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the limit of sequence $\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}$ without using of derivatives and etc. I need to find limit of sequence
$$
\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}\right)
$$
I tried to solve it and stopped here
$$
f... | Let's assume that the limit exists. If we call the limit $s$ then
$s = \sum_1^{\infty}\frac{2n-1}{2^n}$
$\Rightarrow 2s = 1 + \sum_1^{\infty}\frac{2n+1}{2^n} = 1 + \sum_1^{\infty}\frac{2n-1}{2^n} + \sum_1^{\infty}\frac{2}{2^n}$
$\Rightarrow 2s = 1 + s + \sum_0^{\infty}\frac{1}{2^n}$
$\Rightarrow 2s = 1 + s + 2$
$\Right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Solve system:$yz(y+z-x)= a(x+y+z)\\ zx(z+x-y)= b(x+y+z)\\ xy(x+y-z)= c(x+y+z)$ Solve system:
$$yz(y+z-x)= a(x+y+z)\\
zx(z+x-y)= b(x+y+z)\\
xy(x+y-z)= c(x+y+z)$$
$$a, b, c>0$$
I can only solve the system at $a= 2, b= 5, c= 10$. I have tried all things but it's hard with me. Somebody help me?
| From the first equation
$$y z \left(( x + y + z ) - 2 x \right) = a ( x + y + z ) \quad\to\quad
-2x y z = ( a - y z )( x + y + z ) \tag{$\star$}$$
Likewise,
$$-2xyz=(b-zx)(x+y+z) \qquad -2 x y z = ( c - x y )( x + y + z )$$
If, say, $x=0$, then $(a-yz)(y+z) = b(y+z) = c(y+z)=0$; thus, either $y=-z$, or $b=c=0$ and $a=y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Mathematical physics problem. Eletric and Magnetic field. Velocity of propagation of the electromagnetic energy.
If we define the velocity of propagation of the electromagnetic energy, for an arbitrary field in the vacuum, for $\vec{S} = U\vec{v}$ , where $U$ is the density of electromagnetic energy, then
$$\left(1-\f... | This is straightforward. Use $c^2=1/\sqrt{\epsilon_0\mu_0}$.
\begin{align}
\left(1-\frac{v^2}{c^2}\right)U^2&=U^2-\frac{v^2}{c^2}U^2=U^2-\frac{1}{c^2}\left\|\mathbf{S}\right\|^2\\
&=\left[\frac{1}{2}\left(\epsilon_0\left\|\mathbf{E}\right\|^2+\frac{1}{\mu_0}\left\|\mathbf{B}\right\|^2\right)\right]^2-\frac{1}{c^2}\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof Verification: Prove $\sqrt{x}$ is uniformly continuous on [0, $\infty$) Proof:
Fix $\epsilon \gt 0$. We want to find $\delta \gt 0$ such that:
$$|x-a| \lt \delta \Rightarrow |\sqrt{x} - \sqrt{a}| \lt \epsilon$$
$$\phantom{2000i11111}\Rightarrow |\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \epsilon$$
If $|x-a|\lt \delta$ t... | Simply, if $|x - a| < \delta(\epsilon) = \epsilon^2$, then, since $|\sqrt{x} - \sqrt{a}| \leqslant |\sqrt{x} + \sqrt{a}|$, we have
$$|\sqrt{x} - \sqrt{a}|^2 = |\sqrt{x}-\sqrt{a}|\, |\sqrt{x}-\sqrt{a}|\leqslant |\sqrt{x}-\sqrt{a}|\, |\sqrt{x}+\sqrt{a}| = |x - a|< \epsilon^2\\ \implies|\sqrt{x} - \sqrt{a}|< \epsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Let A be a matrix then $A^{50}$ is?
If $$A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\\ \end{bmatrix}$$ then ${A^{50}}$ is?
How to calculate easily? What is the trick behind it? Please tell me.
| Use the fact that:
$$A^2={\begin{pmatrix}1&0&0\\1&1&0\\1&0&1\\ \end{pmatrix}}=I+\begin{pmatrix}0&0&0\\1&0&0\\1&0&0\\ \end{pmatrix}\equiv I+ B$$
Also note that $B^n=0, \forall n>1$, thus using the binomial theorem, and noting that all higher powers of $B$ are dropped:
$$A^{2n}=(I+B)^n=I+nB+ {n\choose 2}B^2+\cdots+ B^n=I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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$y^2+5xy+6x^2-9x-4y=0$ where $x$ and $y$ are integers Need some help solving this for integers $x$ and $y$:
$$
y^2+5xy+6x^2-9x-4y=0
$$
I managed to make something like this:
$$
(y+3x-4)(y+2x)=x\\
(y+3x)(y+2x-3)=y
$$
Find integers for $x$ and $y$ that satisfy the equations above.
But, what do I do next, or is this a bad... | We have $$(y+3x)(y+2x)-(9x+4y)=0$$
so if we put $a=y+3x$ and $b=y+2x$ we get $$ab-a-3b=0\implies b={a\over a-3}$$
so $a-3\mid a$ and thus $a-3\mid 3$ so $a-3\in\{-3,-1,1,3\}$ so $a\in\{0,2,4,6\}$ and corresponding $b\in\{0,-2,4,2 \}$ Now for each pair $(a,b)$ solve for $x$ and $y$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to solve this equation for c? I want to solve the equation
$$
-\frac{a}{2}\left(c+\sqrt{c^2+4}\right)=-\frac{a-1}{2}\left(c-\sqrt{c^2+4}\right)
$$
for $c$, where $a$ is just a constant.
What I get is
$$
\frac{c-\sqrt{c^2+4}}{c+\sqrt{c^2+4}}=\frac{a}{a-1}.
$$
I think there now is some "trick" to solve this for $c$.
| Try to isolate the square root, it is much simpler:
$$-a \sqrt{c^2+4}+a c+\sqrt{c^2+4}-c=a \sqrt{c^2+4}+a c\\
-2 a \sqrt{c^2+4}+\sqrt{c^2+4}-c=0\\
(1-2 a) \sqrt{c^2+4}-c=0\\
\sqrt{c^2+4}=\frac{c}{1-2a}$$
Square both sides and you get:
$$c^2+4=\frac{c^2}{(1-2a)^2}$$
Rearrange that and you have a quadratic equation:
$$(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Using the $N - \varepsilon$ definition to find the limit of a sequence Yesterday I had a post about this, and it cleared a lot up; however, even though I feel like I understand how to go about solving problems like this, I don't seem to get the right answers. For example:
$u_n = \frac{2n+3}{2n+1}$. We know that this se... | I think you are giving yourself too much work here.
Straightforwardly $u_n-1=\frac {2}{2n+1}$ and if $\frac 2{2n+1} \lt \epsilon$ we have $n\gt \frac 1{\epsilon}-\frac 12$
This is reversible, so this will find you a candidate for $N$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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If $\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$, then $\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$
If $$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$$
show that
$$\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$$
where $k$ is an integer such that $2 < k \neq 4$, and where $a$, $b$, $c... | \begin{align}
\text{If }\quad
\frac{b+c}{2k-1}&=\frac{c+a}{2k}=\frac{a+b}{2k+1}
\tag{1}\label{1}
,\\
\text{show that }\quad
\frac{\sin A}{k+1}&=\frac{\sin B}{k}=\frac{\sin C}{k-1}
\tag{2}\label{2}
.
\end{align}
From \eqref{1} we have
by
the rules based on componendo and dividendo,
\begin{align}
\frac{b+c}{2k-1}&=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Maximum Triangle Inscribed So here is an interesting but frustrating problem I have been working on: Inside a square with side length 10, two congruent equilateral triangles are drawn such that they share one side and each has one vertex on a vertex of the square. What is the side length of the largest square that can ... | Start by drawing it out and labeling it like so (sorry, it's not perfect, but good enough):
First find the side length of the equilateral triangle. $G$ is the midpoint of $\overline{FE}$. Let $\overline{GE} = x$. $\overline{DG} = \overline{GB}$, so using basic trigonometry, $\overline{DG} \land \overline{GB}=x\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all solutions of the equation $13[x]+25\{x\}=271$
For a real number $x$, let $[x]$ denote the largest integer $\le x$, and let $\{x \}$ denote $x-[x]$. Find all solutions of the equation:
$$13[x]+25\{x\}=271.$$
I tried to simplify the equation by:
$$13[x]+25(x-[x])=271$$
$$\implies 13[x]+25x-25[x]=271$$
$$\impli... | $\displaystyle \frac{25x-271}{12}=\lfloor x \rfloor$.
\begin{align*}
x-1<\frac{25x-271}{12}&\le x\\
19+\frac{12}{13}< x&\le20+\frac{11}{13}\\
\lfloor x \rfloor&=19 \textrm{ or }20
\end{align*}
If $\lfloor x \rfloor=19$, $\displaystyle x=\frac{12(19)+271}{25}=19.96$.
If $\lfloor x \rfloor=20$, $\displaystyle x=\frac{12(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$
Find $$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$$
without L'Hôpital's rule.
My work:
1) I know that $$\lim_{x\rightarrow0}\frac{\sin x}{x}=1$$
2) Let $x=t-\frac{\pi}{6}$. Then
$$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}=... | Let $x=t+\frac{\pi}{6}$ with $t\to 0$
$$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}=\lim_{t\rightarrow 0}\frac{1-2\sin \left(t+\frac{\pi}{6}\right)}{\cos\left(3t+\frac{\pi}{2}\right)}=\lim_{t\rightarrow 0}\frac{1-\sqrt 3\sin t-\cos t}{-\sin 3t}=\frac{\sqrt 3}{3}$$
indeed
$$\frac{1-\sqrt 3\sin t-\cos t}{-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding value of $\prod_{k=1}^{n-1}\cos\frac{k\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$
Finding value of $$\prod_{k=1}^{n-1}\cos\frac{k\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$$
Try: let $$z^{2n}=1\Rightarrow z=(1)^{\frac{1}{2n}}=e^{\frac{k\pi}{n}}$$
So $$z^{2n}-1= \prod^{2n}_{k=1}(z-e^{\frac{k\pi}{n}})=(z^2-1)\prod^{n-1}_{k=1}(z-... | You are pretty close to derive the answer yourself.
Let $S$ be the product at hand, we have
$$\begin{align}S^2 &= \prod_{k=1}^{n-1} \cos^2\frac{k\pi}{2n} =
\prod_{k=1}^{n-1}\frac12\left(1+\cos\frac{k\pi}{n}\right) =
4^{1-n}\prod_{k=1}^{n-1}\left(2+2\cos\frac{k\pi}{n}\right)\\
&= 4^{1-n}\lim_{z\to-1}\prod_{k=1}^{n-1}\le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Largest cone that can be inscribed in a sphere
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is $\frac{8}{27}$ of the volume of the sphere.
My Attempt
r : radius of the base
h : height of the cone
$$
x=\sqrt{R^2-r^2}\implies h=R+x=R+\sqrt{R^2-r^2}\\
V(r)=\frac{1}{3}\pi r^2h=... | $2R\sqrt {R^2 - r^2} +2R^2 - 3r^2 = 0\\
4R^2 (R^2 - r^2) =(2R^2 - 3r^2)^2\\
4R^4 - 4R^2r^2 = 4R^4 - 12R^2r^2 + 9r^2\\
8R^2 = 9r^2\\
r = \sqrt {\frac 89} R$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\sin(\tfrac{\pi}3) + \tfrac12\sin(\tfrac{2\pi}3) + \tfrac13\sin(\tfrac{3\pi}3)+\dots$ up to infinity Initially, I thought of doing this by first evaluating the function
f(x) = cos(x) + cos(2x) + cos(3x) +.....
and then integrating it. However, I cant seem to find a proper integratable function (if t... | $$\sum^{\infty}_{j=1}\frac{1}{j}\sin \bigg(\frac{j\pi}{3}\bigg)$$
$$=\frac{\sqrt{3}}{2}\bigg[1+\frac{1}{2}-\frac{1}{4}-\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+\cdots\bigg]$$
$$=\frac{\sqrt{3}}{2}\int^{1}_{0}\bigg(1+x-x^3-x^4+x^6+x^7-x^9-x^{10}\cdots\bigg)dx=\frac{\sqrt{3}}{2}\int^{1}_{0}\frac{1+x}{1+x^3}dx$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c... | The determinant of the given matrix is a symmetric polynomial with degree $3$ in the variables $a,b,c$. By the Laplace expansion the determinant equals zero if $a,b$ or $c$ equal zero, hence the determinant is a constant multiple of $abc$. In order to find which multiple, it is enough to evaluate the determinant at $(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Sum to $n$ terms of the given series Find the sum to $n$ terms of the given series:
$$0.3+0.33+0.333+0.3333+\cdots$$
My Attempt:
Let
$$S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$
$$=\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots$$
$$=\frac {3}{10} \left[1+\frac {11}{10}+\frac {... | Let
$$S_n=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$
$$=\underbrace{\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots}_n$$
$$=3\left(\frac {1}{10}+\frac {11}{100}+\frac {111}{1000} + \frac {1111}{10000}+\cdots\right)$$
$$=3\left(\frac {1}{10^1}+\frac {11}{10^2}+\frac {111}{10^3} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\... | Let $g_n = \sqrt{4^n - f_n^2}$
You have noticed that the sequence seem well-behaved at first, and in particular, $f_{n+1}$ and $g_{n+1}$ should look like linear combinations of $f_n$ and $g_n$ except for the issue of the sign of the square root.
$f_{n+1} = \frac 15 (8 f_n + 6 g_n)$ is easy to get, and
$g_{n+1}^2 = 4^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Let $x_1=a>0$ and $x_{n+1}=x_n+\frac{1}{x_n} \forall n\in \mathbb N$. Check whether the following sequence converges or diverges.
Let $x_1=a>0$ and $x_{n+1}=x_n+\frac{1}{x_n} \forall n\in \mathbb N$.
Check whether the following sequence converges or diverges.
When I was in UG my teacher used derivative test for mon... | Extending Clark's answer.
$x_{n+1}^2
=x_n^2+2+\dfrac1{x_n^2}
$
so
$x_{n+1}^2-x_n^2 \ge 2$.
Summing,
$x_{n+1}^2-x_1^2 \ge 2n$
so
$x_{n+1}^2
\ge x_1^2 + 2n
= 2n+a^2
$
so
$x_{n+1}
\ge \sqrt{2n+a^2}
\gt \sqrt{2n}
$.
Therefore
$x_{n+1}
=x_n+\frac{1}{x_n}
\le x_n+\frac{1}{\sqrt{2(n-1)}}
$
or
$x_{n+1}- x_n
\le \frac{1}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
What is the angle between the median and the bisector? In a triangle ABC, what is the angle $\theta$ between the median AM and the bisector AD?
I want a way to know the measure of that angle, given the lenghts of the sides and angles of the triangle. I tried solving the triangle, with the sine and cosine laws, and I ar... | By law of sines:
$$
\frac{\sin CAM}{\sin BAM}=\frac{\sin C}{\sin B}=:\lambda.
$$
Setting $\sin BAM=x$, and solving the resulting equation for cosine of the sum:
$$
\sqrt{1-x^2}\cdot\sqrt{1-\lambda^2x^2}-x\cdot\lambda x=\cos A\\
$$
one obtains:
$$
\sin BAM=\frac{\sin A}{\sqrt{1+\lambda^2+2\lambda\cos A}}
=\frac{\sin A\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Intuitive method of finding probability of getting an onto function from all possible functions from a set to another. The question I was dealing with is as follows:
Let sets $A$ and $B$ have $7$ and $5$ elements, respectively. If one function is selected from all possible defined functions from $A$ to $B$, what is th... | Both your answer and the stated answer of $$\frac{7 \cdot 2!}{3 \cdot 5^6}$$ are incorrect.
Let sets $A$ and $B$ have $7$ and $5$ elements, respectively. If one function is selected from all possible defined functions from $A$ to $B$, what is the probability that it is onto?
You counted the number of functions corre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Partial fractions integral. Compute $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$
Evaluate $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$.
I directly approached this with partial fractions and rewritten $x^6+1=(x^2)^3+1=(x^2+1)(x^4+x^2+1)$.
Therefore the integral is:
$$\int_{0}^{1} \frac {x^4+1}{(x^2+1)(x^4+x^2+1)}dx=-2\int_{0}^{1} \... | Note that $x^6+1=(x^2)^3+1=(x^2+1)(x^4-x^2+1)$ and
use the fact that you are integrating over $[0,1]$,
$$\int_0^1 \frac{x^4+1\pm x^2}{x^6+1}dx =\int_0^1 \frac{x^4-x^2+1}{x^6+1}dx+ \frac{1}{3}\int_0^1 \frac{3x^2}{x^6+1}dx\\=\int_0^1 \frac{1}{x^2+1}dx+ \frac{1}{3}\int_0^1 \frac{1}{t^2+1}dt=\left(1+\frac{1}{3}\right)\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Verify my proof of divisibility of a rational fraction I tried to prove that the following rational fraction can be divided only when n=1,2,5 without using mathematical induction or function, just by using basic method. $$\frac{n^2+2}{2n-1}$$ this expression can be divided only when the remainder of $$\frac{n^2}{2n-1}... | Yet another way to get the answer:
If $2n - 1$ divides $n^2 + 2$ then there is an integer $k$ such that
$n^2 + 2 = k(2n - 1),$
that is,
$$ n^2 - 2kn + (k + 2) = 0.$$
Solving this as a quadratic equation in $n,$
\begin{align}
n &= \frac{2k \pm \sqrt{4k^2 - 4(k + 2)}}{2} \\
&= k \pm \sqrt{k^2 - k - 2)},
\end{align}
whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to calculate probability of $5$ in a row with at least one rare vs $5$ in a row without a single rare? There are $36$ possible outcomes. Event $A$ (later - "rare group") happens in $11$ outcomes. Event $B$ (normal group) in $25$ outcomes ($36-11$). Events $A$ and $B$ cannot occur at the same time - only one of them... | Since the selections are made without replacement, this is actually a hypergeometric distribution.
You correctly calculated that the number of ways no members of the rare group are selected in five trials is
$$\frac{\dbinom{25}{5}}{\dbinom{36}{5}}$$
Therefore, the probability that at least one member of the rare gro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Maximum Value of $a+b+c$
Given that $a+\frac {11}b+\frac c4=20$, what is the maximum value of $a+b+c$? Here, $a,b,c$ are positive integers.
I tried to find the maximum from the AM-GM relationship but failed.
| Notice
$$b > 0 \implies \frac{11}{b} > 0
\implies a + \frac{c}{4} < 20 \iff 4a+c < 80 \implies 4a + c \le 79$$
This leads to
$$b = \frac{11}{20- \left(a + \frac{c}{4}\right)} = \frac{44}{80 - (4a + c)} \le \frac{44}{80-79} = 44$$
Furthermore, $a \ge 1$ implies
$$a + c \le 79 - 3a \le 79 - 3 = 76$$
Combine these, we ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A limit involving $\cos x$ and $x^2$ The question is finding the value of $L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}$ if it exists . I've found the answer using taylor's formula but I'm looking for other solutions maybe using trigonometric identities .
| An answer not using trig identities: rewrite the function as
$$
\begin{align}
&\frac{1}{x^2}\left[1-\left(1-x^2\frac{1-\cos x}{x^2}\right)\left(1-4x^2\frac{1-\cos 2x}{4x^2}\right)\left(1-9x^2\frac{1-\cos 3x}{9x^2}\right)\right]=\\
&\quad=\frac{1}{x^2}[1-(1-x^2f(x))(1-4x^2f(2x))(1-9x^2f(3x))]=\\
&\quad=f(x)+4f(2x)+9f(3x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Solve differential equation $(1+x^2) \frac{dy}{dx} - 2xy = x$ Solve differential equation $$(1+x^2) \frac{dy}{dx} - 2xy = x$$
I simplified it to $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$ \int \frac{1}{1+2y} dy =\int \frac{x}{1+x^2} dx $$
$$\frac{1}{2} \int \frac{2}{1+2y} dy = \frac{1}{2} \int \frac{2x}{1+x^2} dx$$
... | First double both sides to delete the halves, then exponentiate. Thus $k=\pm\exp 2C$, the sign being due to removing the modulus signs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Solve the differential equation: $\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} } $ $$\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} }$$
Simplified it to: $$ \int \frac{1}{\sqrt{1-y^2}} dy = \int \frac{1}{\sqrt{1-x^2}} dx$$
$$\implies\sin^{-1} y = ( \sin^{-1} x + C)$$
$$y = (\sin (\sin^{-1} x) \cos C_1) + \cos(\sin^{-1} x) (\... | Put,
$y$ = $\sin \alpha$ and $x$ = $\sin \beta$
calculate $dy$ and $dx$,
$dy$ = $\cos \alpha$$.d \alpha$ and $dx$ = $\cos \beta$$.d\beta$.
The equation will be something like,
=> $\int \cot\alpha $ $d\alpha$= $\int \cot\beta$ $d\beta$
=> $\log$ |$sin$ $\alpha$| + $c_1$ = $\log$ |$sin$ $\beta$| + $c_2$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Limit of a Function involving tangent function and limits at infinity Determine $$\lim_{x \to \infty}\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$.
Attempt
Let $$y=\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$
Put $\frac{1}{x}=p$.
$$\lim_{p \to 0}\left(\tan{\frac{\pi}{2+p}}\right)^p$$.
We have
$$\lim_{x \... | $$L =\lim_{x \to \infty}\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$
$$\lim_{x \to \infty}g(x) = \lim_{x \to \infty}\left(\tan{\frac{\pi x}{2x+1}}\right) = \lim_{x \to \infty}\left(\frac{2 + 4 x}{\pi}\right)\tag{1}$$
So
$$L = e^{\left(\lim_{x \to \infty}\frac{1}{x}\ln(g(x))\right)}$$
$$\lim_{x \to \infty}\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Shortest distance between ellipsoid and plane Find the shortest distance between the points $A = (x_1,y_1,z_1)$ and $B = (x_2,y_2,z_2)$ if $A$ lies on the plane $x+y+z=2a$ and $B$ lies on the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$
| The plane containing $A$:
$$x+y+z=2a \tag{1}$$
Direction ratio of the normal at $A$ is $1:1:1$
Equation of tangent plane at $B(x_2,y_2,z_2)$:
$$\frac{x_2 x}{a^2}+\frac{y_2 y}{b^2}+\frac{z_2 z}{c^2}=1 \tag{2a}$$
Equating the direction ratios of their normals
$$(x_2,y_2,z_2)=t(a^2,b^2,c^2)$$
As $B$ lies on the ellipsoid,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Minimum splitting field of the polynomial over GF(2) I need to find the minimum splitting field of the polynomial over GF(2): $$x^5+x^4+1.$$
I find that $x^5+x^4+1 = (x + \alpha)^2(x + \alpha + 1)(x + \alpha^2)(x + \alpha^2 + \alpha)$ over GF(8). But I don't know if it's right, because $x =\alpha $ is a multiple root.
... | $x^5+x^4+1=(x^2 + x + 1) (x^3 + x + 1)$ in $GF(2)$.
These factors are irreducible because they have no root in $GF(2)$.
The splitting field of $x^2 + x + 1$ has degree $2$ over $GF(2)$.
The splitting field of $x^3 + x + 1$ has degree $3$ over $GF(2)$.
Therefore, the splitting field of $x^5+x^4+1$ has degree $6$ over $G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Matrices satisfying $(A-B)C=BA^{-1}$ where $A$ is Non singular If $A$,$B$,$C$ are Square matrices satisfying $$(A-B)C=BA^{-1},$$ where $A$ is nonsingular.
Then which is true among these?
*
*$C(A-B)=BA^{-1}$
*$(A-B)C=A^{-1}B$
*$C(A-B)=A^{-1}B$
*$(A-B)^{-1}=C+A^{-1}$
My try:
Since $A$ is invertible,
$$(A-B)CA=B$... | (1) and (2) are incorrect. Take $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $C = \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}$. Verify that the given equation is satisfied.
$$(A-B)C = \begin{pmatrix} ? & -1 \\ ? & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the sum of numbers between 250 and 350 which are divisible by 7?
What is the sum of numbers between $250$ and $350$ which are divisible by $7$?
My attempted solution:
7|250|35
|21 |
-----
40
35
----
5
The 1st number divisible by $7$ is: $(250-5)+7 = 252$.
7|350|50
|35 |
-----
0
The last... | $$
252 + 259 + 266 + 273+\cdots\cdots+336+343+350
$$
Since this is an arithmetic sequence, i.e. the difference between each number and the next is the same in all cases, the average of all the numbers in the list must be the same as the average of the first one and the last one:
$$
\frac{252+350} 2 = 301
$$
Therefore t... | {
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"source": "stackexchange",
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Is this Evaluation of definite Integral correct? To Evaluate $$I=\int_{-1}^{1} \frac{dx}{x^2+x+1+\sqrt{x^4+3x^2+1}}$$
I rationalized the denominator getting as
$$I=\int_{-1}^{1}\frac{x^2+x+1-\sqrt{x^4+3x^2+1}}{2x(x^2+1)}$$ $\implies$
$$I=\int_{-1}^{1}\frac{dx}{2x}+\int_{-1}^{1}\frac{dx}{2(x^2+1)}+\int_{-1}^{1}\frac{\sq... |
Is this Evaluation of definite Integral correct?
No.
Your final result is right, but we have
$$
\int_{-1}^{1}\frac{dx}{x}\ne0,\qquad \int_{-1}^{1}\frac{\sqrt{x^4+3x^2+1}\:dx}{x(x^2+1)}\ne0, \tag1
$$ each integral being divergent, it is the difference of the preceding integrals which vanishes, that is
$$
\int_{-1}^{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $P(x)=x^2 -1$. Find out number of distinct real roots of $P(P(\cdots P(x))\cdots)=0$ where there are $2018$ $P$. Let $P(x)=x^2 -1$. Find out number of distinct real roots of $P(P(\cdots P(x))\cdots)=0$ where there are $2018$ $P$.
My Attempt
Let $Q(x)=P(P(\cdots P(x))\cdots)=0$ where there are $2017$ $P$ is a root o... | Let $a_n$ be the number of roots of the equation with $n P$'s.
Now we have $a_1= 2$. and $a_2 =3$.
For simplicity, denote $P_n$ with the equation of $n P$'s.
For $n \ge 2$, we have : $$ P_n^2 -1 = ((P_{n-2}^2-1)^2-1)^2 -1 = (P_{n-2}^2 -1)^2((P_{n-2}^2-1) -\sqrt{2})((P_{n-2}^2-1) + \sqrt{2}). $$ (Here $P_0(x) =x$).
Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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degree of $\sqrt{2} + \sqrt[3]{3}$ over $\mathbb Q$ How can i find the degree of the minimal polynomial $P \in \mathbb Q[x]$ such that $P(\sqrt{2} + \sqrt[3]{3}) = 0$ ?
Recently i proved that $\mathbb Q[\sqrt{2} + \sqrt{3}] = \mathbb Q[\sqrt{2}, \sqrt{3}]$ using $(\sqrt{2} + \sqrt{3})^{-1} = \sqrt{3} - \sqrt{2}$, so $2... | Hint: It is enough to prove that $\mathbb Q[\sqrt{2} + \sqrt[3]{3}] =\mathbb Q[\sqrt{2},\sqrt[3]{3}]$ because then the facts below imply that the degree of $\sqrt{2} + \sqrt[3]{3}$ is exactly $6$:
*
*$\mathbb Q[\sqrt{2},\sqrt[3]{3}]$ has degree at least $6$
*$\sqrt{2} + \sqrt[3]{3}$ is a root of a polynomial of deg... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove the polynomial $x^2+2x+1$ divides the polynomial $x^{2(2n+1)}+2x^{2n+1}+1$ for every $n\in \mathbb{N}$ I've been trying to prove that the polynomial $x^2+2x+1$ divides the polynomial $x^{2(2n+1)}+2x^{2n+1}+1$ for all $n\in \mathbb{N}$ as follows and reached a "roadblock":
I wrote $x^2+2x+1$ as $(x+1)^2$ and showe... | Take $x^{2m+1}=a$
Now, $x^{2(2m+1)}+2x^{2m+1}+1=a^2+2a+1$
$\Rightarrow$$(a+1)^2$ $\Rightarrow$$(x^{2m+1}+1)^2$
So, $x+1|x^{2m+1}+1$ $\Rightarrow$$(x+1)^2|x^{2(2m+1)}+2x^{2m+1}+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding value of $\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$
Finding value of $$\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$$
Solution I tried:
$$(1+x)^{10}=1+\binom{10}{1}x+\binom{10}{2}x^2+\cdots +\binom{10}{10}x^{10}$$
$$(x+1)^{10}=x^{10}+\binom{10}{1}x^9+\binom{10}{2}x^7+\cdots +\binom{10}{10}$$
I did not find h... | To complete the method sketched in the comments.
The answer is $\binom {13}5$.
To see this, suppose we are choosing $5$ values from $\{1,\cdots, 13\}$. Order the choices as $n_1<\cdots <n_5$. Let $n_3=k+1$. Clearly $2≤k≤10$. Given a choice of such an $n_3$ we see that we are asked to choose $2$ values from $\{1,\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the flux of $F=(x+3y^5, y+10xz, z-xy),$ $ S$ the closed hemisphere bounded by $x^2+y^2+z^2=1, z\ge0$. Calculate the flux across the surface $S$ of $F=(x+3y^5, y+10xz, z-xy),$ $ S$ the hemisphere bounded by $x^2+y^2+z^2=1, z\ge0$.
I have done $n=(x,y,z)$
thus $\iint F\cdot n dS= \iint(x+3y^5, y+10xz, z-xy)\cd... | $$div F=(F_1)'_x+(F_2)'_y+(F_3)'_z=3$$
$$\int\int\int_V3dxdydz=3.\frac {2}{3}\pi 1^3=2\pi $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If three sides of an acute triangle is $3$ cm, $4$ cm, and $x$ cm. What are the possible values of $x$? Problem: If three sides of an acute triangle is 3 cm, 4 cm, and $x$ cm. What are the possible values of $x$?
2 of the choices: (A) $1<x<5$, and (B) $0<x<7$
Solution: According to triangle theorems, 2 sides of a trian... | Let the angles opposite the sides of lengths $x,3,4$ be $\theta,\omega, \phi$ respectively. Then by the Cosine Rule we obtain the relations: $$x^2=3^2+4^2-2\cdot3\cdot4\cos \theta,$$ $$3^2=x^2+4^2-2\cdot x\cdot4\cos \omega$$ and $$4^2=x^2+3^2-2\cdot x\cdot3\cos \phi$$ for $0<\cos \theta, \cos \omega,\cos \phi<1$.
From ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Integrate $\int \frac {dx}{1+\sin x}$ Integrate $\int \dfrac {dx}{1+\sin x}$
My attempt:
$$\begin{align}&=\int \dfrac {dx}{1+\sin x}\\
&=\int{\dfrac{dx}{1+\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1+\tan ^2\left( \dfrac{x}{2} \right)}}}\\
&=\int{\dfrac{1+\tan ^2\left( \dfrac{x}{2} \right)}{\left( 1+\tan \left( \dfrac{x... | Hint:$$\int\frac1{1+\sin x}\,\mathrm dx=\int\frac{1-\sin x}{\cos^2x}\,\mathrm dx=\int\frac1{\cos^2x}\,\mathrm dx-\int\frac{\sin x}{\cos^2x}\,\mathrm dx.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find $\int\arcsin(\sqrt{x})dx$
Find $\displaystyle\int\arcsin(\sqrt{x})dx$
My Attempt
Put $y=\sqrt{x}\implies dy=\frac{1}{2\sqrt{x}}dx\implies dx=2ydx$
$$
\int\arcsin(\sqrt{x})dx=2\int \arcsin(y)\,y\,dy=2\bigg[\frac{y^2}{2}\arcsin(y)-\int\frac{1}{\sqrt{1-y^2}}\frac{y^2}{2}dy\bigg]\\
=y^2\arcsin(y)-\int\frac{y^2}{\sqr... | Hint. Note that
$$-\int\frac{y^2}{\sqrt{1-y^2}}\,dy=\int\frac{1-y^2}{\sqrt{1-y^2}}\,dy-\int\frac{dy}{\sqrt{1-y^2}}=\int\sqrt{1-y^2}\,dy-\arcsin(y).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Evaluating $\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}.$ I tried to use the series for $\sin \pi x$ and maybe find something related to $\zeta(3)$, but didn't work. I'm guessing this integral needs more than the little calculus that I know.
\begin{equation}
\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \... | $\displaystyle f(a):=\int\limits_0^1 x e^{ax}dx = \frac{1+e^a(a-1)}{a^2}$
$\displaystyle g(a):=\int\limits_0^1 x^2 e^{ax}dx = \frac{-2+e^a(a^2-2a+2)}{a^3}$
$\displaystyle \int\limits_0^1 \frac{x^2-x}{\sin(\pi x)}dx = i2\int\limits_0^1\frac{x^2-x}{e^{i\pi x}-e^{-i\pi x}}dx = i2\sum\limits_{k=0}^\infty \int\limits_0^1 (x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 4
} |
Derivative of a trigonometric function $\arctan$ Will someone help me explain this example please?
$$y=\arctan \dfrac{\sqrt{1-\cos (x)}}{\sqrt{1+\cos (x)} }$$
Solution:
$$\sin^2 (x/2)=\dfrac{1-\cos (x)}{2}\Rightarrow \sqrt{2}\sin (x/2)=\sqrt{1-\cos (x)}$$
$$\cos^2 (x/2)=\dfrac{1+\cos (x)}{2}\Rightarrow \sqrt{2} \cos... | Without the preliminary simplifications.
Let $$y=\tan^{-1}(u) \qquad \text{with} \qquad u=\dfrac{\sqrt{1-\cos (x)}}{\sqrt{1+\cos (x)} } $$
$$\frac{dy}{dx}=\frac 1 {1+u^2}\frac{du}{dx}=\cos ^2\left(\frac{x}{2}\right)\,\frac{du}{dx}$$ Now $$\log(u)=\frac 12 \log(1-\cos(x))-\frac 12 \log(1+\cos(x))$$ Differentiate both si... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How do I know when to take ordering into account?
I'm quite confused as to why in (c) the ordering matters but not in (b)
I kind of understand in (a) ordering doesn't really matter since all the numbers are the same therefore switching places doesn't really make any difference.
However, for (c) the solution is showin... | It helps to distinguish between the dice. Suppose the dice are, respectively, blue, green, and red. Then an outcome can be represented by the ordered triple $(b, g, r)$, where $b, g, r \in \{1, 2, 3, 4, 5, 6\}$.
I roll three six-sided fair dice. What is the probability that all three dice show the same number?
Sin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int e^{-3x} \cos^3x\,dx$
Evaluate $\int e^{-3x}\cos^3x\,dx$
My Attempt
\begin{align}
& \int e^{-3x} \cos^3x\,dx=\cos^3x \cdot \frac{e^{-3x}}{-3}-\int3\cos^2x\sin x \cdot \frac{e^{-3x}}{-3} \, dx \\
= {} &\frac{-1}{3}\cos^3x \cdot e^{-3x}+\int\cos^2x\sin x \cdot e^{-3x} \, dx\\
= {} & \frac{-1}{3}\cos^3x\cd... | Hint: try to use the trigonometric identity
$$\cos^3(x)= \frac{1}{4}(3\cos(x)+\cos(3x))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Solving recurrent equation $a_1=3; a_2=1; a_n = a_{n-1} + 2a_{n-2}+2$ I have the following recurrent equation
$
a_1=3\\a_2=1\\a_n=a_{n-1}+2a_{n-2}+2
$
and I cannot seem to be able to solve it correctly. According to wolframalpha and my math textbook (that contains results but not the actual process), the result should ... | Well I would do like this. Since for all $n$ we have:
$$
a_n-a_{n-1}-2a_{n-2}=2 $$
we have also (change $n$ to $n+1$) $$a_{n+1}-a_{n}-2a_{n-1}=2$$
so if we substract these two we get: $$a_{n+1}-2a_n-a_{n-1}+2a_{n-2}=0$$
so zeroes of characteristic equation $x^3-2x^2-x+2=0$ are $1,2,-1$ so $$a_n = a\cdot (-1)^n+b\cdot 2... | {
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Find the marginal probability density function The random vector $[\,X \,\,\, Y \,]'$ has probability density function
$f_{X,Y} (x,y) = ke^{-2x^2-3xy-\frac{9}{2}y^2}$, where $k$ is some constant
Find $k.$
Find the marginal probability density functions of $X$ and $Y.$
I know for it to be a valid pdf its integral from ... | A standard form of the bivariate normal density with expected value $(0,0)$ is this:
$$
f(x,y) = \frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}}
\exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{x^2}{\sigma_X^2} + \frac{y^2}{\sigma_Y^2} - \frac{2\rho xy}{\sigma_X \sigma_Y} \right] \right)
$$
The exponent is this:
$$
-\f... | {
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"url": "https://math.stackexchange.com/questions/2799788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$ Recently I have been reading physics book and saw interesting equation, like this:
$$\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
But I still don't understand how to get the right part of the equation from left... | We can prove this in a reverse way
Lets start with $$\frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
$$=\frac{1/a}{\sqrt{\frac{a^2-b^2}{a^2}}}$$
$$=\frac{1/a}{\frac{\sqrt{a^2-b^2}}{a}}$$
$$=\frac{\frac{1}{a}(a)}{\sqrt{a^2-b^2}}$$
$$=\frac{1}{\sqrt {a^2 - b^2}}$$
Hence,$$\frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}=\frac{1}{\sqrt ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral $I=\int_0^1 \frac{\ln x\arctan(ax)}{1+b^2x^2}dx$ Greetings I am trying to evaluate $$I=\int_0^1 \frac{\ln x\arctan(ax)}{1+b^2x^2}dx$$ Where $a$ and $b$ are positive numbers. My try was to derivate the integral with respect to $a$ in order to get: $$I'(a)=\int_0^1 \frac{\ln x}{(1+a^2x^2)(1+b^2x^2)}dx=\frac{1}{... | For one, you're missing an $x$ in the numerator when you differentiating $\arctan ax$.
Let $F(a)$ denote our integral and differentiating with respect to the parameter gives
$$\begin{align*}F'(a) & =\int\limits_0^1 dx\,\frac {x\log x}{(1+a^2x^2)(1+b^2x^2)}\\ & =\frac {a^2}{a^2-b^2}\int\limits_0^1dx\,\frac {x\log x}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Show that $(9-5\sqrt 3)(2-\sqrt 2) $ is not a square in $\mathbb Q (\sqrt 2, \sqrt 3) $
Show that $(9-5\sqrt 3)(2-\sqrt 2) $ is not a square in $\mathbb Q (\sqrt 2, \sqrt 3) $; equivalently, prove that $\mathbb Q (\sqrt 2, \sqrt 3, u)$ is of degree $2$ over $\mathbb Q (\sqrt 2, \sqrt 3) $ where $u $ is a number such t... | Here is a more "hands-on" (and systematic) approach. Suppose that
$$u^2=(2-\sqrt{2})(9-\sqrt{3})\tag{1}$$
for some $u\in{\mathbb Q}(\sqrt{2},\sqrt{3})$. You can write $u=v+w\sqrt{3}$
with $v,w\in{\mathbb Q}(\sqrt{2})$. Expanding in (1) yields
$$
(v^2+3w^2)+(2vw)\sqrt{3}=9(2-\sqrt{2})-(2-\sqrt{2})\sqrt{3} \tag{2}
$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding roots using $\displaystyle\int_0^1(1+\cos^8x)(ax^2+bx+c)dx=\int_0^2(1+\cos^8x)(ax^2+bx+c)dx $ I have the following relation-
$$\displaystyle\int_0^1(1+\cos^8x)(ax^2+bx+c)dx=\int_0^2(1+\cos^8x)(ax^2+bx+c)dx $$
$$\text{then what is the interval in which the root of the equation}\space ax^2+bx+c=0\space\text{l... | $$\int_0^1(1+\cos^8x)(ax^2+bx+c)dx=\int_0^2(1+\cos^8x)(ax^2+bx+c)dx$$
$$\Leftrightarrow$$
$$\int_0^1(1+\cos^8x)(ax^2+bx+c)dx - \bigg(\int_0^1 (1+\cos^8x)(ax^2+bx+c)dx $$
$$+ \int_1^2 (1+\cos^8x)(ax^2+bx+c)dx \bigg)= 0$$
$$=$$
$$\int_1^2(1+\cos^8x)(ax^2+bx+c)dx=0$$
Observe that :
$$1+\cos^8(x) >0 \; \forall \; x\in \ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Obliques asymptotes of a logarithmic and rational Function with absolute value. I'm trying to find the obliques asymptotes of the following function:
$$f(x)=|x-1| \log\Big(\sqrt{x^2 +3x+3}-x-1\Big)$$
Its domain should be this: $D(f)=(-\infty, +\infty)$;
I discussed the absolute values: $$|x-1|=\left\{
\begin{array}{c}... | We have
\begin{align}
\log\bigl(\sqrt{x^2 +3\,x+3}-x-1\bigr)-\log\Bigl(\frac{1}{2}\Bigr)&=
\log\Bigl(\frac{x+2}{\sqrt{x^2 +3\,x+3}+x+1}\Bigr)-\log\Bigl(\frac{1}{2}\Bigr)\\
&=\log\Bigl(\frac{2\,x+4}{\sqrt{x^2 +3\,x+3}+x+1}\Bigr)\\
&=\log\Bigl(1+\frac{x+3-\sqrt{x^2 +3\,x+3}}{\sqrt{x^2 +3\,x+3}+x+1}\Bigr)\\
&=\log\Bigl(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Laurent series of $f(z) = \frac{5}{z^2 + (2i - 1)z - 2i}$ Let $f: \mathbb{C} \setminus \left\{ 1, -2i \right\} \rightarrow \mathbb{C}$ with $$f(z) = \frac{5}{z^2 + (2i - 1)z - 2i}$$
Determine the Laurent series of $f$ in the annuli $\left\{ z \in \mathbb{C} : 0 < |z + 2i| < \sqrt{5} \right\}$.
I think I didn't understa... | By a standard fraction decomposition, one has
$$
f(z) = \frac{5}{z^2 + (2i - 1)z - 2i}=\frac{1-2i}{z+2i}+\frac{2i-1}{1-z}
$$wich rewrites
$$
\frac{1-2i}{z+2i}+\frac{2i-1}{1-z}=\frac{1-2i}{z+2i}+\frac{2i-1}{1+2i}\cdot\frac{1}{1-\frac{z+2i}{1+2i}}=\frac{1-2i}{1+2i}\cdot \frac1Z+\frac{2i-1}{1+2i}\cdot\frac{1}{1-Z}
$$with
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $x+y$, if integers $x$ and $y$ satisfy the equation $y+1/x=25/3$ Find $x+y$, if $x$ and $y$ are integers and satisfy the equation $y+1/x=25/3$
so I got to the answer by placing $3$ in the $x$ cause it looked like a fraction and $8$ was left for $y$,
My question is: Is there any other/better way to solve this alge... | We can solve it algebraicly by considering divisiblity by 3 (see below).
But we can also justify why your intuition yields the only possible answer.
No $\frac {25}3 = 8\frac 13$ and $8< 8\frac 13 < 8+1$.
If $x > 1$ then $y < y+ \frac 1x< y + 1$.
So $y = 8; \frac 1x = \frac 13; x=3$.
If $x < -1$ then $y-1 < (y-1) + (1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n-1}I tried using mathematical induction to prove this, but the problem I faced was that there are a lot of numbers between $\frac{1}{2^k-1}$ and $\frac{1}{2^{k+1}-1}$. Is it possible to prove this with induction or is there a better method?
| $$
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^n-1}=1+\left(\frac{1}{2^1}+\frac{1}{2^2-1}\right)+\left(\frac{1}{2^2}+\cdots+\frac{1}{2^3-1}\right)+\cdots+\left(\frac{1}{2^{n-1}}+\cdots+\frac{1}{2^n-1}\right)<n,
$$
since
$$
\frac{1}{2^{k-1}}+\cdots+\frac{1}{2^k-1}<2^{k-1}\cdot\frac{1}{2^{k-1}}=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Cool way of finding $\cos\left(\frac{\pi}{5}\right)$ while trying to solve a problem, I stumbled upon a way of finding $\cos\left(\frac{\pi}{5}\right)$ using identities and the cubic formula. Is it possible to find other values of sine or cosine in a similar way ?
Consider
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\... | Not so cool. The following may be cooler.
Let $\Delta ABC$ be a triangle with $\angle BAC=36^o$, $\angle ABC=\angle ACB=72^o.$ $BD$ bisects $\angle ABC$. $BE\perp AC$. You would find that $AD=BD=BC$ and $\Delta ABC \sim \Delta BDC.$ Thus, as the figure shows, we may obtain $$\frac{x}{2y}=\frac{x+2y}{x}.$$ Thus $$\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Common tangent(s) to two parabolas without calculus We have the parabolas: $y = 2x^{2} + 2x + 1$ and $y = 2x^{2} - 2x - 1$.
Finding the common tangents with calculus is as straight forward as just solving a system of equations with the derivatives of both parabolas and $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$; but how abo... | A line $y=kx+m$ is a tangent to the parabola $y=ax^2+bx+c$ iff the quadratic equation
$$
kx+m=ax^2+bx+c
$$
has exactly one solution, which makes the discriminant to be zero. In our case
$$
kx+m=2x^2+2x+1\iff 2x^2+(2-k)x+1-m=0,\\
kx+m=2x^2-2x-1\iff 2x^2-(2+k)x-1-m=0
$$
should each have a unique solution $x$, that is th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Splitting up an infinite sum I am playing with the following example trying to determine if $\frac{1}{n}-\frac{1}{\sqrt{n}}$ is in $\ell^2$.
Here is what I have,
\begin{align}
\sum\limits_{n=1}^\infty \left( \frac{1}{n}-\frac{1}{\sqrt{n}}\right)^2 &= \sum\limits_{n=1}^\infty \left(\frac{\sqrt{n}-n}{n\sqrt{n}}\right)^2\... | It is not in $\ell^2$:
\begin{align}
\sum_{n=1}^\infty\frac{n-2n\sqrt n+n^2}{n^3}&=\sum_{n=1}^\infty\frac{1-2\sqrt n+n}{n^2} \\
&= \sum_{n=1}^\infty\frac{(\sqrt{n}-1)^2}{n^2} \\
&\ge \sum_{n=1}^3\frac{(\sqrt{n}-1)^2}{n^2} + \sum_{n=4}^\infty\frac{\left(\frac12\sqrt{n}\right)^2}{n^2} \\
&= \sum_{n=1}^3\frac{(\sqrt{n}-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding $f(x)$ from the functional equation $(e^x-1)f(2x)= \left(e^{2x}-1\right)f(x)$ and $f'(0) =1 $ Let $f: \mathbb R \to \mathbb R$ be a non constant continuous function such that $(e^x-1)f(2x)= \left(e^{2x}-1\right)f(x)$. If $f'(0) =1 $, then $\lim_{x\to 0}\left(\dfrac {f(x)}{x}\right)^{\frac 1x}= ? $
I am trying h... | This solution only assumes $f$ is differentiable at $0$.
Fix $x\ne 0$ and notice that
$$f(x)\left(e^{\frac{x}2} - 1\right) = f\left(2\cdot \frac{x}2\right)\left(e^{\frac{x}2} - 1\right) = f\left(\frac{x}2\right)(e^x-1)$$
so inductively we get
$$f\left(\frac{x}{2^n}\right)(e^x-1) = f(x)\left(e^{\frac{x}{2^n}} - 1\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify: $\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$ Simplify:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$$
So what I've tried was:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=\frac{1+\cos^2x-\sin^2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=$$
$$=\... | The computation can be much shorter if one uses the relevant duplication/linearisation formulæ:
*
*$\displaystyle \frac{1+\cos 2x}{1-\cos x}=\frac{2\cos^2x}{1-\cos x}=\frac{\cos^2x}{\,\smash[b]{\cfrac{1-\cos x}2}\,}=\frac{\cos^2x}{\,\sin^2\dfrac x2\,}$.
$\phantom{fg}$
*
*$\displaystyle \frac{4\cos^2\dfrac x2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability that all three shirts were worn Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.
What is the probability that he wears all three shirts after 6 day... | I will try to replicate @Mustafa's answer in a different way.
Guess there is $6$ white tshirts, $3$ green, and $2$ red. For this person to have picked one atleast from each of all of them during 6 days it sums to $\sum\binom{1}{6}^i\binom{1}{3}^j\binom{1}{2}^k$ for $i$ $j$ $k$ strictly positive and $i+j+k<=6$.
Taking c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$3^n-1$ is divisible by $4 \implies n $ is even What is the easiest ways to prove this: $3^n-1$ is divisible by $4 \implies n $ is even? Moreover, how would I figure out that $n$ must be even if I didn't know the result?
My approach is this: suppose $n=2k+1$. Consider $3^{2k+1}-3$. If we show that it is divisible by $4... | We can use the identity
$a^n - b^n = (a - b) \displaystyle \sum_0^{n - 1} a^{n - 1 - k}b^k: \tag 0$
$3^n - 1 = (3^1 - 1^1) \displaystyle \sum_0^{n - 1} 3^{n - 1 - k} 1^k = 2 \sum_0^{n - 1} 3^{n - 1 - k}; \tag 1$
now if
$4 = 2^2 \mid 3^n - 1 = 2 \displaystyle \sum_0^{n - 1} 3^{n - 1 - k}, \tag 2$
then
$2 \mid \displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
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Solve for $y$ in the equation $x = \frac{3y^2 + 14y + 16}{6y^2 + 24y + 24}$ As part of a larger calculation, I came across $x = \frac{3y^2 + 14y + 16}{6y^2 + 24y + 24}$, which I now have to solve for $y$. My initial idea, besides a failed attempt to use the general quadratic formula, was, incorrectly:
\begin{align}
\do... | $$ (3y+8)(y+2) = 3 y^2 + 14 y + 16 $$
$$ 6 (y+2)^2 = $$
$$ x = \frac{3y+8}{6y+12} $$
This is a Mobius transformation
$$ y = \frac{12x - 8}{-6x+3} $$
Audience Request:
If we have constants $a,b,c,d$ with $ad-bc \neq 0,$ and
$$ y = \frac{ax+b}{cx+d} $$
when $cx+d \neq 0,$ then
$$ x = \frac{dy - b}{-cy + a} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving $a_{n+1}^2=a_n·a_{n+2}+(-1)^n$ for Fibonacci sequence $\{a_n\}$ Let $a_1,a_2,a_3,...,a_n$ be the sequence of Fibonacci numbers.
Then we are required to prove by induction that $$a_{n+1}^2=a_n \cdot a_{n+2} + (-1)^n$$
My Attempt:
Initially putting a few values like $n=1,2 \: \text{or}\ 3$ makes it certain that w... | Induction step $n\to n+1$
$$\begin{align}a_{n+1}^2-a_n \cdot a_{n+2} &= a_{n+1}^2-a_n \cdot (a_{n+1}+a_n) \\
& =a_{n+1}(a_{n+1}-a_n)-a_n^2
\\&= a_{n+1}a_{n-1}-a_n^2
\\&= (a_n^2-a_{n+1}a_{n-1})
\\&= -(-1)^{n-1}\\
&= (-1)^n\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Where is the mistake in my reasoning? I have a statement that says:
If $a^2 + b^2 + c^2 = 2$ and $(a + b + c)(1 + ab + bc + ac) = 3^2$
What is the value of $( a + b + c )$ ?
My reasoning was:
$a^2 + b^2 + c^2 = 2$, rewritten as:
*
*$(a + b + c)^2 = 2 + 2ab + 2ac + 2bc$
Since, $(a + b + c)(1 + ab + bc + ac) = 3^... | Suppose $a^2+b^2+c^2=2$ and $(a+b+c)(1+ab+bc+ac)=k$. Then, setting $s=a+b+c$ and $q=ab+bc+ac$, we get
$$
s^2=2+2q=2(1+q),\qquad s(1+q)=k
$$
Therefore $1+q=s^2/2$ and
$$
s^3=2k
$$
so that $s=\sqrt[3]{2k}$. If $s$ has to be $4$, then $2k=64$, that is, $k=32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 0
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How does one construct general forms that certain variables in an equation must take? Srinivasa Ramanujan was one of the greatest mathematicians of all time $-$ the greatest in the $20^\text{th}$ century. One day, he stumbled across the equation $$\rm3^3+4^3+5^3=6^3\tag1$$ and only days later, he was able to discover t... | (OP) enquiry about derivation of the Ramanujan Identity. Derivation is given below:
Assume that below equation is true:
$\big(3p^2+mp-5\big)^3+\big(4p^2-np+6\big)^3+\big(5p^2-mp-3\big)^3=\big(6p^2-np+4\big)^3\tag1$
Let:
$u=3p^2+mp-5$
$v=4p^2-np+6$
$w=5p^2-5p-3$
$z=6p^2-np+4$
We have identity:
$(a+b+c)^3=a^3+b^3+c^3+3(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is $x$ if $\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$? I need to find $x$, given that
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
I simplified this to $x^4+14x^3+105x^2+68x-188=0$. According to Symbolab, that is not correct. That's why I'm goint to write out my attempt so you can point out where my mistake is.
My attempt... | What you did is fine. Now, you can use the rational root theorem in order to find the roots $1$ and $-2$. Since your polynomial is $(x-1)(x+2)(x^2+13x+94)$, there are no more real roots. Note however that you still must check whether or not $-2$ and $1$ are solutions of the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Ellipse: Most direct algebraic demonstration that $\left|x\right|$ is maximal when $y=0$ Yes, I know this is the equation of an ellipse.
This is the starting point of the problem:
$$
r_1+r_2=2a,
$$
where
$$r_1\equiv\sqrt{(x+c)^2+y^2}; r_2\equiv\sqrt{(x-c)^2+y^2}.$$
I want to establish certain properties prior to deriv... | This is a constrained optimization problem that can be attacked directly with the method of Lagrange multipliers. Form the Lagrangian $x-\lambda(r_1+r_2-2a)$ and differentiate, producing the equations $$1-\lambda\left({x+c \over r_1}+{x-c \over r_2}\right) = 0 \\
-\lambda\left({y\over r_1}+{y\over r_2}\right) = 0.$$ Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generating function of a recurrent relation Problem: Find the generating function for the recurrent relation:
$$f_n=2f_{n-1}+\frac 12 f_{n-2},$$
where
$$f_0=f_1=1.$$
My idea was to first find a few of the beginning values and then try to make into some sum of an infinite series, but I had no luck. Any ideas?
|
We denote with $F(x)=\sum_{k=0}^\infty f_n x^n$ a generating function for the recurrence relation
\begin{align*}
f_n&=2f_{n-1}+\frac{1}{2}f_{n-2}\qquad\qquad n\geq 2\tag{1}\\
f_0&=1\\
f_1&=1
\end{align*}
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
W... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit with Epsilon - Delta method Prove using the $\epsilon - \delta$ definition of limits that $\lim_{x\to3} \frac{5}{4x-11} = 5$.
I know how the setup should be given $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that $|x-3| \lt \delta$ and $|\frac{5}{4x-11} - 5| \lt \epsilon$ but I can't do the computation to ... | Let $\varepsilon > 0$ and define $\delta = \min\left\{\frac{\varepsilon}{4(5+\varepsilon)}, \frac14\right\}$. For $|x-3| < \delta$ we have
$$\left|5 - \frac{5}{4x-11}\right| = \left|\frac{20x-60}{4x-11}\right| \le \frac{20\left|x-3\right|}{\left|4x-11\right|} \le \frac{20\left|x-3\right|}{1 - \left|4x-12\right|} = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m... | If the numbers are $x, x+1, x+2, x+3$.
Let $\frac m2 = x+1.5$ be the midpoint of the four consecutive integers, so that the integers are $\frac {m-3}2, \frac {m-1}2, \frac {m+1}2, \frac {m+3}2$. (Note: $m$ is odd and $\frac m2$ is not an integer.)
So $x(x+1)(x+2)(x+3) + 1 = $
$\frac {(m-3)(m+3)(m-1)(m+1)}{16} + 1=$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Find the set of primes $p$ which $6$ is a quadratic residue $\mod p$ Since $6$ is not prime (law of quadratic reciprocity could have been used), how does one find the set of primes $p$ for which $6$ is a quadratic residue $\pmod p$? I noticed that $6$ is a quadratic residue $\pmod p$ for the following $p$ up to $100: 2... | Use the fact that Legendre symbol is completely multiplicative:
$$1=\left(\frac{6}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{3}{p}\right)$$
which means either
$$\left(\frac{2}{p}\right)=\left(\frac{3}{p}\right)=1$$
or
$$\left(\frac{2}{p}\right)=\left(\frac{3}{p}\right)=-1$$
and (same link)
$${\displaystyle \left({\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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how to classify critical points for a 2 variable function For each of the following functions, find and classify all critical points. [That is, use the second-derivative test to deduce whether each critical point is a local max, a local min, or a saddle.]
$f(x,y) = (x+y)(1-xy)$
my solution:
I first equated gradie... | With
$f(x, y) = (x + y)(1 - xy) \tag 1$
and
$\nabla f = (f_x, f_y), \tag 2$
we see that
$f_x = 1 - xy + (x + y)(-y) = 1 - xy - xy - y^2 = 1 - 2xy - y^2, \tag 3$
and likewise
$f_y = 1 - xy + (x + y)(-x) = 1 - xy - x^2 - xy = 1 - 2xy - x^2; \tag 4$
at critical points of $f(x, y)$, we have
$\nabla f(x, y) = 0, \tag 5$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Minimal area of a right triangle with inradius $1$ I got this question, solved it, then forgot how I solved it.
What is the minimal area of a right triangle with inradius $1$?
My attempt:
$r=\frac{a+b-c}2$, so $a+b=c+2$
$a^2+b^2=c^2$
This gives $ab=2(c+1)$
I remember using the AM-GM to prove that equality held, thus ... | $$1=r=\frac{a+b-c}{2}=\frac{a+b-\sqrt{a^2+b^2}}{2}.$$
We'll prove that
$$S\geq3+2\sqrt2.$$
Indeed, we need to prove that:
$$\frac{ab}{2}\geq(3+2\sqrt2)\left(\frac{a+b-\sqrt{a^2+b^2}}{2}\right)^2$$ or
$$(a+b+\sqrt{a^2+b^2})^2\geq2(3+2\sqrt2)ab,$$
which is true by AM-GM:
$$(a+b+\sqrt{a^2+b^2})^2\geq(2\sqrt{ab}+\sqrt{2ab}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On the property $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$ for complex numbers Suppose we have complex numbers, $x$ and $y$. My question is: we know, from the triangle inequality, that $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$. It makes sense, geometrical... | \begin{align*}
&|x + y| = |x| + |y| \\
\implies \, &|x + y|^2 = (|x| + |y|)^2 \\
\implies \, &|x + y|^2 = |x|^2 + |y|^2 + 2|x||y| \\
\implies \, &(x + y)\overline{(x + y)} = x\overline{x} + y\overline{y} + 2|x||y| \\
\implies \, &x\overline{x} + y\overline{y} + x\overline{y} + y\overline{x} = x\overline{x} + y\overline... | {
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"url": "https://math.stackexchange.com/questions/2841302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many integral values of $k$ such that $2x^3+3x^2+6x+k=0$ has exactly three real roots How many integral values of $k$ such that $2x^3+3x^2+6x+k=0$ has exactly three real roots. I am unable to see how I'd start this question. A small hint, or the entire solution, both will be highly appreciated!
| Since this is a cubic polynomial.
By using Vieta's formula we have $a+b+c=-\dfrac{3}{2}$ and $ab+bc+ca=3$. Now squaring $a+b+c$ gives $a^2+b^2+c^2+2(ab+bc+ca)=\dfrac{9}{4}$, so $a^2+b^2+c^2=-\dfrac{15}{4}$ which is always impossible if the roots are real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Finding the value of $\tan B$
In the triangle $\triangle ABC$
$$\cot A=\dfrac{-3}{4}$$
$$\sin A \cos B - \cos A \cos B = 1 $$
*
*Find the value of $\tan B$
This $\sin A \cos B - \cos A \cos B = 1 $ reminds me of sum and difference formulas in a particular way that made me draw a triangle. However, I couldn't see an... | $$\begin{align*}
1&=\sin A\cos B-\cos A\cos B\\[1ex]
&=\cos A\cos B(\tan A-1)\\[1ex]
&=\cos A\cos B\left(\frac1{\cot A}-1\right)\\[1ex]
&=-\frac73\cos A\cos B
\end{align*}$$
$$\implies\cos B=-\frac37\sec A\quad(*)$$
Then by the Pythagorean identities, we have
$$\begin{align*}
\sin^2B&=1-\cos^2B\\[1ex]
&=1-\left(-\frac3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the given limit: $\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$ Evaluate the given limit
$$\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$
My Attempt :
$$=\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$
$$=\lim_{x\to 0} \dfrac {x(\tan (2x)-2\tan (x))}{1... | With equivalents, there are less calculations:
First rewrite the expression as $\;\dfrac{x(\tan 2x-2\tan x)}{(1-\cos 2x)^2}$.
Now,
*
*it's standard that $\;1-\cos u\sim_0\dfrac{u^2}2$, so $\;(1-\cos 2x)^2\sim_0(2x)^2 $.
*Taylor's formula at order $3$ for the tangent is $\;\tan u=u+\dfrac{u^3}3+o(u^3)$, so
$$\tan 2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Compute $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
Evaluate $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
My attempt:
$I=\int_0^{2\pi}\frac 1{\sin^4x+\cos^4x}dx=\int_0^{2\pi}\frac 1{(\sin^2x+\cos^2x)^2-2\sin^2(2x)}dx=\int_0^{2\pi}\frac {1}{1-2\sin^2(2x)}dx=\frac 12\int_0^{4\pi}\frac 1{1-2\sin^2(x)}dx=\frac 12 \int_0^{4\... | Under $x\to\tan x\to x-\frac1x$, one has
\begin{eqnarray}
&&\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx\\
&=&4\int_0^{\pi/2} \frac 1{\sin^4x+\cos^4x}dx\\
&=&4\int_0^{\pi/2} \frac {\sec^2x}{1+\tan^4x}\sec^2xdx\\
&=&4\int_0^{\infty} \frac{1+x^2}{1+x^4}dx\\
&=&4\int_0^{\infty} \frac{1+\frac1{x^2}}{x^2+\frac1{x^2}}dx\\
&=&4\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 0
} |
How to evaluate $\lim _{x\to 0}\frac{5-5\cos\left(2x\right)+\sin\left(4x\right)}{x}$ without using L'Hospital's rule? I need to evaluate the following limit without using L'Hopital's rule:
$$\lim _{x\to 0}\left(\frac{5-5\cos\left(2x\right)+\sin\left(4x\right)}{x}\right)$$
I thought the best way was to separate it in tw... | Just another way using Taylor series.
$$\cos(x)=1-\frac{x^2}{2}+O\left(x^4\right)\implies \cos(2x)=1-2 x^2+O\left(x^4\right)$$
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)\implies \sin(4x)=4 x-\frac{32 x^3}{3}+O\left(x^4\right)$$
$$5-5\cos\left(2x\right)+\sin\left(4x\right)=4 x+10 x^2-\frac{32 x^3}{3}+O\left(x^4\right)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\int \sqrt {3 \tan^2 \theta - 1} d \theta$ Evaluate $I=\int \sqrt {3 \tan^2 \theta - 1} d \theta$
My attempt
$\tan \theta = t, $ then $I = \int \frac{\sqrt {3t^2-1}}{1+t^2} dt $
Now integrating by parts,
I = $\sqrt {3t^2-1} \tan^{-1} t- \int( \frac{6t}{2\sqrt {(3t^2-1)}} \tan^{-1} t) dt$
Now i am struck...... | Hints :
$$=\int \frac {(2\tan \theta\sec^2\theta)(\sqrt {3\tan ^2\theta-1})}{(2\tan \theta\sec^2\theta)} d\theta$$
Let $$\tan^2\theta=\frac {x^2+1}{3}$$ hence $$2\tan \theta\sec^2\theta=2x/3$$
And then change the integral to $$\sqrt 3\int \frac {x^2}{(x^2+4)\sqrt {x^2+1}}dx= \sqrt 3\int \frac {1}{\sqrt {x^2+1}}dx-4\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find and classify the singularities of $f(z) = \frac{1}{e^{z^2}-1}$ I am trying to find and classify the singularities of $f(z) = \frac{1}{e^{z^2}-1}$. I'd also like to find the residue for any poles. So far, I have found that the singularities are of the form $z=\sqrt{2\pi k i}$ for $k \in \mathbb{Z}$.
I believe that... | Here we have three cases: $z=0$, $e^{z^2}=1$ and singularity at infinity as $z=\infty$. The series
\begin{align}
f(z)
&=\dfrac{1}{e^{z^2}-1}\\
&=\dfrac{1}{z^2+\frac{1}{2!}z^4+\frac{1}{3!}z^6+\cdots}\\
&=\dfrac{1}{z^2}\cdot\dfrac{1}{1+\frac{1}{2!}z^2+\frac{1}{3!}z^4+\cdots}\\
&=\dfrac{1}{z^2}\left(1-\frac{1}{2!}z^2+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$ Show that $ f(x)=x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$
I tried to simplify it by putting $x^5=y$
It simplifies the polynomial but I cannot put it in the case of the divisor.
So I assumed that $x^2+1$ is a divisor of $f(x)$
Then examine the assumpt... | There is no need for synthetic division. Just evaluate $x^{20} + x^{15} + x^{10} + x^5 $ at $x=\pm i$:
$$
(\pm i)^{20} + (\pm i)^{15} + (\pm i)^{10} + (\pm i)^5
=
1 + \mp i - 1 + \pm i
= 0
$$
Therefore, $x\pm i$ divides $x^{20} + x^{15} + x^{10} + x^5 $ and so does their product, because their are coprime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
Is mod of exponent allowed? Lately I was studying about Modular Arithmetic and the way modulus is used to calculate large numbers. What caught my attention was calculating powers with modulus. It's generally that we calculate the mod of the base and then proceed with our further calculations. Now I was thinking, is it ... | I'm going to guess what the intent of your question is. Two columns of numbers below are surmounted by arrows pointing downward. In the first of those columns, going from each horizontal row to the next, one multiplies by $3$ and reduces modulo $11$ at each step. In the second of those two columns, one reduces modulo ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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.