Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Trigonometric inequality $\sin^{4}x+\cos^{4}x>a$, for $a\in [0,1]$ Solve the inequality:
$\sin^{4}x+\cos^{4}x>a$, for $a\in [0,1]$
My attempt:
$(\sin^{2} x+\cos^{2}x)^2-2\sin^{2}x\cos^2{x}>a$
$\Rightarrow 1-2\sin^{2}x\cos^{2}x>a$
$\Rightarrow 1-2\sin^{2}x(1-\sin^{2}x)>a$
$\Rightarrow 2\sin^{4}x-2\sin^{2}x+1>a$
$t:=\... | Using $\cos2A=1-2\sin^2A=2\cos^2A-1,$
$$\sin^4x+\cos^4x=\dfrac{(1-\cos2x)^2+(1+\cos2x)^2}4=\dfrac{2+2\cos^22x}4=\dfrac{2+1+\cos4x}4$$
Now $-1\le\cos4x\le1$ for real $x$
Alternatively,
$$\sin^4x+\cos^4x=1-2\sin^2x\cos^2x=1-\dfrac{\sin^22x}2=1-\dfrac{1-\cos4x}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that I cannot factor a cubic polynomial with Algebra alone How do I prove a given polynomial is not factor-able using only radicals, basic arithmetic operations, and real, rational numbers? For example, how can I prove that
$$0=x^3-3x+1$$
Has no solution with the given restraints?
Persistently, I want an alg... | So we have that $$x^3-3x+1=0 \tag1$$
We can rewrite this as:
$$\begin{align} & x^3+3x^2+3x+1=3x^2+6x \\
\implies & (x+1)^3=3x(x+2)=3\{(x+1)-1\}\{(x+1)+1\}=3(x+1)^2+2(x+1)-1 \\
\implies & (x+1)^3-3(x+1)^2-2(x+1)+1=0 \\
\implies & (x+1)^3-3(x+1)^2+3(x+1)-1=5(x+1)-2 \\
\implies & (x+1-1)^3=5(x+1)-2 \\
\implies & x^3=5x+3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Show that the 6 lines can be represented by the following equation: $(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$
Question: Figure $3$ shows six lines passing through the origin. The lines are separated by equal angles. Some
exact values of $\tan(t)$ are given in Table $1$.
$(i)$ Show that the lines ... | Part (i)
See diagram above.
Equations are:
$$\begin{align}\color{red}{x=\pm y \qquad\Rightarrow x^2-y^2=0}\\
\color{blue}{x=\pm y\tan\frac{\pi}{12} \qquad \Rightarrow x^2-y^2\tan^2\frac{\pi}{12}=0}\\
\color{green}{y=\pm x\tan\frac{\pi}{12} \qquad\Rightarrow x^2-y^2\big / \tan^2\frac{\pi}{12}=0}
\end{align}$$
Multi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is the sum of the roots of $x^3+9=12x$ zero? I found the roots of the equation to be
$$x=3,\frac{\sqrt{21}}{2}-\frac{3}{2},-\frac{\sqrt{21}}{2}-\frac{3}{2}$$
Algebraically, why is the sum of these roots equal to zero?
| Consider the following cubic form:
$$ax^2+bx^2+cx+d=0$$
For your equation:
$$1\cdot x^3+0\cdot x^2-12\cdot x+9=0$$
Using Vieta's formulas, the sum of roots is equal to $-\frac{b}{a}$.
$$-\frac{0}{1}=0$$
Thus the sum of roots is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A difficult question on representing a square by a sum of 4 (or 5 ) squares The question above is not about the decomposition of a square $$m^2 = a^2 + b^2 + c^2 + d^2 $$ as a sum of 4 or more squares. We know it is always possible and we also have algorithms to do that.
My question is about going backward, that is i... | Not uniquely.
Since for instance, given $a=3, b=5, c=11$
$18^2= 3^2+5^2+11^2+13^2$
$78^2= 3^2+5^2+11^2+77^2$
As we can write $(m+d)(m-d) = a^2+b^2+c^2$
there are as many solutions for $m,d$ as there factorisations of $a^2+b^2+c^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Linear programming vertex proof I want to prove that $x$ is vertex of $\{x\in \mathbb{R}^n\ \colon Ax\leqslant b, \ x\geqslant 0\}$ iff $(x,b-Ax)$ is the vertex of $\{(x,u)\in \mathbb{R}^{n+m}\ \colon Ax+u=b, \ x\geqslant 0, \ u\geqslant 0\}$. Here $A$ is a $m\times n$ matrix.
Definition is: $x \in X$ is called a vert... | Denote the first polyhedral as $P_1$ and the second polyhedral be $P_2$.
Part 1:
Suppose $x \in P_1$ is not a vertex of $P_1$, then $\exists y,z \in P_1$ such that $$x=\lambda y+(1-\lambda)z$$
check that $\begin{pmatrix} x \\ b-Ax \end{pmatrix}\in P_2$
and we have $$\begin{pmatrix} x \\ b-Ax \end{pmatrix}=\lambda \begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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quadratic congruence using Newton-Raphson Method from calculus Question:
Find two solutions of $x^2 + 2x + 2 \equiv 0 \bmod 5^3$
My attempt:
*
*rewrite the equation or simplify (optional step): $x^2 + 2x + 2 = (x+1)^2 + 1 \equiv 0 \bmod 5^3$
*$x = 1$ then: $f(1) = 5$ and $5 \equiv 0 \bmod 5$
*$f'(x) = 2(x+1)$ ... | Use the quadratic formula.
$x_{1,2} = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{-2 \pm \sqrt{4-8}}{2} \pmod{125} = \dfrac{-2 \pm \sqrt{121}}{2} \pmod{125} = \dfrac{-2 \pm 11}{2} \pmod{125} = 2^{-1} \times (9, -13) \pmod{125} = 2^{-1} \times (9, 112) \pmod{125} = 63 \times (9, 112) \pmod{125}= (67, 56)$.
Note that the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Remainder in polynomial division The remainder when $x^{50}$ is divided by $(x-3)(x+2)$ is of the form $ax + b$. Find the units digit of $a$.
I tried to tackle the problem using the polynomial remainder theorem but got stuck as the divisor is a quadratic expression.
| Hint
$x^{50}=(x-3)(x+2)q(x)+ax+b$ (Division algorithm)
Now, take $x=3$ and $x=-2$ to get two equations. Two variables and two equations. Hope you get it.
This is the complete problem. (Use the hint and try yourself first)
Taking $x=3$,
$3^{50}=3a+b,$
Taking $x=-2$
$(-2)^{50}=2^{50}=-2a+b$
Subtracting, we get
$3^{50}-2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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What is $z^4=-4$ using the exponential form method, and with the answers written as $z = x + yi$? I know how to algebraically solve equations like this and find square roots of complex numbers, but how would you solve equations like that with the $n$-th power of $z$ being greater than $2$?
| In exponential form, $z^4 = -4 = 4 \text{ cis } \pi$, where $\text{cis } \theta$ is shorthand for $e^{i\theta} = \cos \theta + i \sin \theta$. Then $z_0 = \sqrt{2} \text{ cis } \frac{\pi}{4}$ (since $\left(\sqrt{2}\right)^4 = 4$), and $z_k$ = $\sqrt{2} \text{ cis } \frac{(1+2k)\pi}{4}$, for $k = 1, 2, 3$.
In rectangul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}$ implies $a_n/n^2$ converges? Let $a_0=\pi$, $a_1=\pi^2$, $a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}$ How can we prove that $a_n/n^2$ converges?
It is easy to see that
$$a_n-a_1=\sum_{k=1}^{n-1}\frac{2a_{k-1}}{k+1}\geq 2a_0\sum_{k=1}^{n-1}\frac{1}{k+1}\to\infty.$$
Then how to do?
| Let $(b_n)=(\frac {a_n} {n^2})$, we have that
\begin{align}
b_{n+1}-b_n&=\frac {n^2a_{n+1}-(n+1)^2a_n}{n^2(n+1)^2}\\
&=\frac {n^2(a_{n+1}-a_n)-(2n+1)a_n}{n^2(n+1)^2}\\
&=\frac {\frac {2n^2}{n+1}a_{n-1}-(2n+1)a_n}{n^2(n+1)^2}\\
&=\frac {2n^2a_{n-1}-(2n+1)(n+1)a_n}{n^2(n+1)^3}\\
&=\frac {2n^2(a_{n-1}-a_n)-(3n+1)a_n}{n^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Then sum of digits in $M\times N$
If $\bf{M = \underbrace{777777777777777777777}_{99-times}}$ and $\bf{N = \underbrace{999999999999999999999}_{77-times}},$ Then sum of digits in $M\times N$
$\bf{My\; Try::}$ We can write $$M = 7(1+10+10^2+\cdots \cdots +10^{98}) = \frac{7}{9}(10^{99}-1)$$
and write $$N = 9(1+10+10^2... | Note that $N=10^{77}-1$, hence $MN=$
$\underbrace{7\dots7}_{99}\cdot(10^{77}-1)=$
$\underbrace{7\dots7}_{99}\cdot10^{77}-\underbrace{7\dots7}_{99}=$
$\underbrace{7\dots7}_{99}\underbrace{0\dots0}_{77}-\underbrace{7\dots7}_{99}=$
$\underbrace{7\dots7}_{76}6\underbrace{9\dots9}_{22}\underbrace{2\dots2}_{76}3$
Hence the d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Limit at negative infinity of a function with a radical $$\lim_{x \to -∞}{\sqrt{x^2 -x} + x}.$$
First I rationalized, to get
$$\frac{x^2-x-x^2}{\sqrt{x^2-x}-x}$$
Then I wanted to factor the whole thing by the largest power(x) in the denominator, to get:
$$\frac{-x/x}{1/x(\sqrt{x^2-x}-x)}$$
After simplifying:
$$\frac{-1... | $$\lim _{ x\to -∞ }{ \sqrt { x^{ 2 }-x } +x } =\lim _{ x\to -∞ }{ \frac { \left( \sqrt { x^{ 2 }-x } +x \right) \left( \sqrt { x^{ 2 }-x } -x \right) }{ \left( \sqrt { x^{ 2 }-x } -x \right) } } =\lim _{ x\to -∞ }{ \frac { -x }{ \left( \sqrt { x^{ 2 }-x } -x \right) } } =\\ =\lim _{ x\to -∞ }{ \frac { -x }{ \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Compute the sum of binomial probability I have a sum that is $$\sum_{k=\lfloor{\frac{n}{2}+1}\rfloor}^n\binom{n}{k}p^n$$
I try $n=3$ and $n=4$, I got
$$\sum_{k=\lfloor{\frac{3}{2}+1}\rfloor}^3\binom{n}{k}p^3=4p^3$$
$$\sum_{k=\lfloor{\frac{4}{2}+1}\rfloor}^4\binom{n}{k}p^4=5p^4$$
How do I do it for all $n$?
| Hint. Note that $\lfloor{\frac{n}{2}\pm 1}\rfloor=\lfloor{\frac{n}{2}}\rfloor\pm 1$, and $\binom{n}{k}=\binom{n}{n-k}$ for $k=0,\dots,n$.
Hence
$$2^n=\sum_{k=0}^n\binom{n}{k}=\sum_{k=0}^{\lfloor{\frac{n}{2}}\rfloor-1}\binom{n}{k}+\binom{n}{\lfloor{\frac{n}{2}}\rfloor}+\sum_{k=\lfloor{\frac{n}{2}}\rfloor+1}^n\binom{n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Does series $(\cosh\frac{1}{n} - 1)^p$ converge? Given: series $\sum^\infty_{n=1}(\cosh\frac{1}{n} - 1)^p$.
Question: Consider different $p$ and find out if it converges.
I tried to represent $\cosh$ as $\frac{e^x - e^{-x}}{2}$. What I got as $n$-th member:
$$\left(\frac{\sqrt[n]{e} + \frac{1}{\sqrt[n]{e}} - 2}{2}\rig... | I would directly use the Taylor polynomial for $\cosh x$ and some basic comparisons. We have
$$\cosh x-1=\frac{x^2}2+R_2(x)=\frac{x^2}2+O(x^3)$$
We thus have that $(\frac1{2n^2}+cn^{-3})^p\le(\cosh\frac1n-1)^p\le(\frac1{2n^2}+Cn^{-3})^p$ for some constants $c,C$. By comparison with $n^{-2p}$, this is summable if and on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the volume of the regions enclosed by $z = x^2+y^2-2$ and $z = 30-x^2-y^2$
Find the volume of the regions enclosed by $z = x^2+y^2-2$ and $z = 30-x^2-y^2$
I set up a triple integral with the bounds of the inmost as $x^2 + y^2 - 2$ to $30 - x^2 - y^2$. The two outer integrals both had the bounds from $-4$ to $4$.... | Set $z = z$ in the two equations to find the circle of intersection:
$x^2 + y^2 - 2 = 30 - x^2 - y^2\\
2x^2 +2y^2 = 32\\
x^2+ y^2 = 16$
$\iiint \;dz\;dy\;dx\\
\iint z\;dy\;dx$
$\int_{-4}^{4}\int_{-\sqrt{16-x^2}}^{\sqrt {16-x^2}} 32 - 2x^2- 2y^2\;dy\;dx$
Now we could keep cranking through that in Cartesian, but I think ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to relate the $|z^3+z^{-3}|$ with $|z+z^{-1}|$? I got stuck on this seemingly simple question:
If $z$ is a complex number satisfying $|z^3+z^{-3}| \le 2$, then the maximum possible value of $|z+z^{-1}|$ is:
(A) $2$
(B) $2^{1/3}$
(C) $2\sqrt 2$
(D) $1$
Using the AM-GM inequality, I showed that $|z|^3 + |z|... | $(z+z^{-1})^3=z^3 + 3z + 3z^{-1}+z^{-3}$ therefore $|z+z^{-1}|^3 \le |z^3 + z^{-3}| + 3 |z+z^{-1}| \le 2 + 3|z+z^{-1}|$ and using notation $|z+z^{-1}| = x$ we have:
$x^3 \le 2 + 3x$ equivalent $(x-2)(x+1)^2 \le 0$ therefore $|z+z^{-1}| \le 2 $ is the answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Value of $\sin(\pi/18)$ by nested radicals I recently asked a question about how to find value of $\sin(\pi/18)$ and I understand that there is no expression for $\sin(\pi/18)$ that uses the ordinary arithmetical operations. but I know that we have exact value by nested radicals. I want to know how we get this value ?
... |
Theorem 1: For $a_1=\sqrt a$, $a_2=\sqrt{a-\sqrt{a}}, a_3=\sqrt{a-\sqrt{a+\sqrt a}},\ldots,$
We have $a_n$ as$$\lim_{n\rightarrow\infty}a_n=\frac {A-1}6+\frac 23\sqrt{4a+A}\sin\left(\frac 13\arctan\frac {2A+1}{3\sqrt3}\right)\tag1$$where $A=\sqrt{4a-7}$.
Proof: I present to you the proof for the theorem.
Start with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
... | Modular inverses can be rotely computed by the extended Euclidean algorithm, as well as other less-known methods methods that are sometimes simpler for small numbers. A few are below.
Fiirst we consider Gauss's algorithm, which scales the (top & bottom) of the fraction to make the bottom smaller when reduced mod $41$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 0
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Practice test - simplifying expressions another question I am stuck on in a practice test.
The question is $$\frac{5^{2011} - 5^{2009} +24}{ 5^{2009} +1}$$
Can you cancel out the $5^{2009}$ or not?
| Hint:
$$
\begin{align}
\frac{5^{2011} - 5^{2009} +24}{ 5^{2009} +1} & = \frac{5^2(5^{2009}+1) - 5^2 - (5^{2009} + 1) + 1 +24}{ 5^{2009} +1} \\
& = 25 - 1 + \frac{\;\;\cdots\;\;}{5^{2009} +1}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Calculating ...5(5+4(4+3(3+2(2+1(1))))) I've been puzzling over this question for a while now, and I've finally decided to turn to the StackExchange community in order to get an answer.
How would one determine the value of the expression ...5(5+4(4+3(3+2(2+1(1)))))) to n, assuming that in this case n = 5?
Cheers!
| This can be written as:
$$5\cdot 5 + 5\cdot 4\cdot 4 + 5\cdot 4\cdot 3\cdot3 + 5\cdot 4\cdot 3\cdot 2\cdot 2+5!\cdot 1\\
=\frac{5!}{4!}\cdot 5 + \frac{5!}{3!}\cdot 4 + \frac{5!}{2!}\cdot 3 + \frac{5!}{1!}\cdot 2+\frac{5!}{0!}\cdot 1$$
Then the general formula is:
$$\sum_{k=0}^{n-1}\frac{n!}{k!}(k+1) = n!\sum_{k=0}^{n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Four distinct integers a,b,c,d are in A.P. If $a^2+b^2+c^2=d$, then find$ a+b+c+d$. Four distinct integers a,b,c,d are in arithmetic progression. If $a^2+b^2+c^2=d$, then find $a+b+c+d$.
My attempt:-
| As you did, let $a=A-x,b=A$.
Then, we have
$$2x^2\color{red}{-}2x+3A^2-A=0,$$
i.e.
$$x=\frac{1\pm\sqrt{-6A^2+2A+1}}{2}$$
Since $x$ has to be an integer, let $-6A^2+2A+1=m^2$ to have
$$6m^2+\left(6A-1\right)^2=7$$
So, $7-6m^2\ge 0\implies m=0,\pm 1$ giving $(m,A)=(\pm 1,0)$.
Since $a,b,c,d$ are distinct,
$$a+b+c+d=4A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Upper and lower bound of $\sum\limits_{n=1}^{m}\sin(nx)$? According to a post in Desmos, the $\sum\limits_{n=1}^{m}\sin(nx)$ is squeezed between $\frac{1}{2}\cot\left(\frac{x}{4}\right)$ and $-\frac{1}{2}\tan\left(\frac{x}{4}\right)$ where
$$\frac{1}{2}\cot\left(\frac{x}{4}\right)<\sum_{n=1}^{m}\sin(nx)<-\frac{1}{2}\ta... | Note that $\sin(a)\sin(b)=\frac{1}{2}\left(\cos(a-b)-\cos(a+b)\right)$.
So your partial sum is also:
$$f_n(x)=\sum_{m=1}^{n}\sin mx =\frac{1}{2} \frac{\cos\frac{x}{2}-\cos\frac{2n+1}{2}x}{\sin\frac{x}2}$$
When $x\in (0,2\pi)$, this means that:
$$f_n(x)\leq \frac{1}{2}\frac{\cos\frac{x}2+1}{\sin\frac{x}{2}}=\frac{1}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving System of Linear Equations with LU Decomposition of $4 \times 3$ matrix The following is all confirmed to be true:
Matrix A =
$
\begin{bmatrix}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{bmatrix}
$
U =
$
\begin{bmatrix}
-1 & 2 ... | Multiplying the first row of $L$ with the first column of $U$ gives us $-1$, which is not the $(1,1)$ entry of $A$. Hence, you have made a mistake in computation of LU decomposition.
There seems to be a missing permutation matrix being involved in your computation.
Your procedure to solve the linear system upon comput... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Necklace problem with Burnside's lemma
How many necklaces can be made with two red beads, two green
beads, and two violet beads?
I'm trying to solve this with Burnside's Lemma however I'm stuck. Here's what I've got so far:
Let $S=\{$necklace of length 6 with 2 read, 2 green and 2 violet$\}$, $|S|=\frac{6!}{2!2... | I upvoted the first answer but I would like to show how to compute the
cycle index $Z(D_6)$ of the dihedral group $D_6$ and apply the
Burnside lemma and the Polya Enumeration Theorem to this problem.
Observe that the OEIS uses the convention of refering to rotational
symmetry as a necklace and as a brac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $\lim\limits_{n\to\infty}\sqrt [n] {n^{2}+n+1}=1$ Proof. Note that $1\leq \sqrt [n] {n^2+n+1}\leq \sqrt [n] {3n^{2}}$. But,
$\lim\limits_{n\to\infty}\sqrt [n] {3n^{2}} =1$. Therefore, by the squeeze lemma, we have $\lim\limits_{n\to\infty} \sqrt [n]{n^2+n+1} =1$
My question is: How did $\lim\limits_{n\to\inf... | We can write $\sqrt[n]{3n^2} = (\sqrt{3}n)^\frac{2}{n} = e^\frac{2\ln(\sqrt{3}n)}{n}$. Note that for $n > 1$ we have $0 < \ln(n) < n$. If $n > 1$ then $\sqrt{\sqrt{3}n} > 1$ (since $\sqrt{3}$ > 1). Thus we have for $n > 1$
\begin{align}
0 < \ln(\sqrt{\sqrt{3}n}) < \sqrt{\sqrt{3}n}\\
0 < 2\ln(\sqrt{\sqrt{3}n}) < 2\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now... | Had never heard of the box method before I saw you use it!
When you got to the part of factoring $x^2 + 4x + 3$ I would go and find the roots of it, because with the roots one can also factor the polynomial.
Upon finding that $-1$ and $-3$ are the roots, I would know $x^2 + 4x + 3 = (x - (-3))(x - (-1)) = (x + 1)(x + 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 0
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How should this fraction involving powers be solved
$$\sqrt { 5 } \cdot { \left( \frac { 5 }{ 4 } \right) }^{ \frac { 1 }{ 2 } }$$
All I know is that this can be written as
$\sqrt { 5 } \cdot { \left( \frac { 5 }{ { 2 }^{ 2 } } \right) }^{ \frac { 1 }{ 2 } }$
Any Ideas?
| $\sqrt { 5 } { \left( \frac { \sqrt5 }{ \sqrt4 } \right)}=$${\left( \frac { \sqrt {5*5} }{ 2 } \right)}$=${\left( \frac { \sqrt {25} }{ 2 } \right)}$=$\frac 52$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\sum_{n=1}^\infty \frac{i^n}{n}$ converge? $$\sum_{n=1}^\infty \frac{i^n}{n}$$
I'm not sure how to approach this series to find if it converges/diverges because of the $i^n$. I tried using the comparison test comparing it with $\frac{1}{n}$ and the ratio test, but didn't get anywhere. How would you approach this and l... | Since $i^n = i^{n+4}$,
each 4 consecutive terms
is of the form
$\begin{array}\\
\frac{i}{4m+1}+\frac{-1}{4m+2}+\frac{-i}{4m+3}+\frac{1}{4m+4}
&=\frac{i}{4m+1}+\frac{-1}{4m+2}+\frac{-i}{4m+3}+\frac{1}{4m+4}\\
&=i(\frac{1}{4m+1}-\frac{1}{4m+3})-\frac{1}{4m+2}+\frac{1}{4m+4}\\
&=i(\frac{(4m+3)-(4m+1)}{(4m+1)(4m+3)})-(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Help me solve a combinatorial problem Consider a ordered set consisting of $N$ integers. It is called beautiful if for every integer $x$ in the set, either $x-1$ or $x+1$ is present in the set.
How many beautiful sets can be created using the numbers $1$ to $M$ of size $K$ for some arbitrary $M$ and $K$?
The number o... | We want to count the number of strings of length $M+1$ consisting of $M-K+1$ atoms that look like
$$
\underbrace{\ 0\ }_1,\underbrace{110}_{x^2},\underbrace{1110}_{x^3},\underbrace{11110}_{x^4},\dots
$$
That is, there are $M-K+1$ zeros and $K$ ones. Since the atoms all end with a zero, we've added one to the number of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How many minutes does Train $1$ take for the entire trip?
Two trains simultaneously depart towards each other on one route of Washington's Metro line. Train $1$ departs from Point A on one end and Train $2$ departs from Point B on the other end. At the instant the trains pass each other, it takes Train $1$ another $16... | So you let the speed of train $1$ be $s_1$ and train $2$ be $s_2$. Let us call the distance between $A$ and $B$ as $x$. Let the distance between their common meeting point $C$ and $A$ be $y$, so that $BC$ is then $x-y$.
Then,
1) It is clear that train $2$ takes $7$ minutes less than train $1$. So one equation would be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the range of $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$, for $a$, $b$, $c$ the sides of a triangle
If $a$, $b$, and $c$ are the three sides of a triangle, then
$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$
lies in what interval?
| The problem is Nesbitt's inequality. (Source: A.M. Nesbitt, Problem 15114, Educational Times (2) 3, (1903), 37-38.)
Lower Bound: Let's say $S= \dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b} $. $S+3= \dfrac{a}{b+c}+1 + \dfrac{b}{c+a}+1 + \dfrac{c}{a+b}+1=(a+b+c)\left( \dfrac{1}{b+c} + \dfrac{1}{c+a} + \dfrac{1}{a+b}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Bernoulli numbers. Today I read a short article about Bernoulli numbers. This article give the way to calculate Bernoulli numbers.
$\displaystyle \frac{x}{e^{x}-1} = f(x)$
$\displaystyle f(x)(e^{x}-1) = x = \sum_{n = 0}^{\infty}B_{n} \frac{x^{n}}{n!} \sum_{k = 1}^{\infty} \frac{x^{k}}{k!} = \sum_{n=1}^{\infty}x^{n} \su... | In slow motion
$$
\begin{array}{l}
\sum\limits_{0\, \le \,n} {B_{\,n} \frac{{x^{\,n} }}{{n!}}} \sum\limits_{1\, \le \,k} {\frac{{x^{\,k} }}{{k!}}} = \sum\limits_{\begin{array}{*{20}c}
{0\, \le \,n} \\
{1\, \le \,k} \\
\end{array}} {B_{\,n} \frac{{x^{\,n + k} }}{{n!k!}}} = \sum\limits_{\begin{array}{*{20}c}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Can the antiderivate of $\frac{1}{(x^2+1)^2}$ be calculated without using the iteration formula? To calculate the antiderivate of $$\frac{1}{(x^2+1)^2}$$ , we can either use the iteration formula reducing the exercise to the integral of $\ \frac{1}{x^2+1}\ $ or we can use $$(\frac{x}{x^2+1})'=\frac{1-x^2}{(x^2+1)^2}$$ ... | We have
$$\int\frac{dx}{(1+x^2)^2}=\int \frac{1+x^2-x^2}{(1+x^2)^2}dx=$$
$$\arctan(x)+\frac{1}{2}\int x\frac{-2x}{(1+x^2)^2}dx=$$
$$\arctan(x)+\frac{1}{2}\left(\left[x \frac{1}{1+x^2}\right]-\int \frac{1}{1+x^2}\right)=$$
$$\frac{1}{2}\arctan(x)+\frac{x}{ 2(1+x^2) }+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve for $x$ in $\log_4{(x+4)} \le \log_2{(2x+5)}$
I would like to solve $$\log_4{(x+4)} \le \log_2{(2x+5)}$$ for $x$.
I did:
$$\log_4{(x+4)} \le \log_2{(2x+5)} \Leftrightarrow \frac{\log(x+4)}{\log(4)} \le \frac{\log(2x+5)}{\log(2)}\\ \Leftrightarrow \frac{\log(x+4)}{2\log(2)} \le \frac{\log(2x+5)}{\log(2)} \Leftr... | Hint: Use that $\log(2)>0$ and that $t\mapsto \log(t)$ and $t\mapsto e^t$ are monotonic functions to continue:
\begin{align}
\frac{1}{2} \cdot \frac{\log(x+4)}{\log(2)} \le \frac{\log(2x+5)}{\log(2)} &\iff \frac{1}{2}\log(x+4) \le \log(2x+5) \\ &\iff \log(\sqrt{x+4}) \le \log(2x+5) \\ &\iff \sqrt{x+4} \leq 2x+5\\ &\iff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Calculate the exact value of $\sin\frac{11\pi}{8}$ Calculate the exact value of $\sin\frac{11\pi}{8}$.
The formula $\sin^2x=\frac12(1–\cos2x)$ may be helpful.
I was thinking of using the Angle-Sum and -Difference Identity:
$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$
For instance: $\alpha=\frac{20\pi}{... | As $\dfrac{11\pi}8=\pi+\dfrac{3\pi}8$ which lies in the third Quadrant,
$\sin<0,\cos <0,\tan>0$
Now $\tan\dfrac{11\pi}8=\cdots=\tan\dfrac{3\pi}8$
Using $\tan A=\dfrac{1-\cos2A}{\sin2A},\tan\dfrac{3\pi}8=\dfrac{1-\cos\dfrac{3\pi}4}{\sin\dfrac{3\pi}4}$
$\sin\dfrac{3\pi}4=\sin\left(\pi-\dfrac\pi4\right)=\sin\dfrac\pi4=?$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Number of distinct sequences of length 10, containing at least 5 consecutive As or at least 5 consecutive Bs Im stuck on this question, asking for the number of distinct sequences of length 10, which contain at least 5 consecutive As or 5 consecutive Bs (for example ABABBBBBBA should be counted, as should ABBBBAAAAA). ... | The generating function for the number of strings with at most $4$ As in a row and $4$ Bs in a row is
$$
\begin{align}
g(x)
&=\overbrace{\ \frac{1-x^5}{1-x}\ }^\text{$0$-$4$ A}\overbrace{\frac{1\vphantom{x^5}}{1-\underbrace{\frac{x-x^5}{1-x}}_\text{$1$-$4$ B}\underbrace{\frac{x-x^5}{1-x}}_\text{$1$-$4$ A}}}^\text{$0$+}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find $2^{n-1}\prod_{k = 1}^{n-1}{\left(\cos\theta - \cos\left( \frac{k\pi}{n} \right)\right)}$ Evaluate:
$$2^{n-1} \left(\cos\theta - \cos\left( \frac{\pi}{n} \right)\right)\left(\cos\theta - \cos\left( \frac{\pi}{n} \right)\right)...\left(\cos\theta - \cos\left( \frac{(n-1)\pi}{n} \right)\right)$$
The answer is
\beg... | I will use multiple angle formula to solve the problem:
$$\sin(mx) = 2^{m-1}\prod_{k=0}^{m-1} \sin\left(x+\frac{k\pi}{m}\right)$$
Now, to solve our problem,
\begin{align}
&2^{n-1} \prod_{k=1}^{n-1} \left( \cos \theta - \cos \left( \frac{k\pi}{n}\right) \right)\\ & = 2^{n-1} \prod_{k=1}^{n-1} (-2) \sin \left(\frac{\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2038523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The sequence $a_n=\frac{2n+1}{(n-1)^2}$ decreases monotonically How could we show that the sequence $a_n=\frac{2n+1}{(n-1)^2}$ decreases monotonically?
When we take the quotient $\frac{a_n}{a_{n+1}}$ we get $\frac{n^2(2n+1)}{(n-1)^2(2n+3)}$. Howcan we conclude that this quotient is $\geq 1$ ?
| $\frac{n^2(2n+1)}{(n-1)^2(2n+3)}>1 \iff 2n^3+n^2>(n^2-2n+1)(2n+3) \iff 2n^3+n^2>2n^3+3n^2-4n^2-6n+2n+3 \iff 2n^2+4n-3>0 \iff (\sqrt{2}n+\sqrt{2})^2-5>0 \iff (\sqrt{2}n+\sqrt{2})^2>5$
Since the last inequality true and we have only biconditionals, it follows that the first inequality, which is precisely what we want, i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Show $f(x,y)=(\frac{x}{|x|+|y|},\frac{y}{|x|+|y|})$ it bjective and continuous and its inverse is continuous I'm working through Renzo's Math 490 Introduction to Topology, http://www.math.colostate.edu/~renzo/teaching/Topology10/Notes.pdf, and on page 16, Example 1.8.4, it states:
that a topologist cannot tell the ... | Injectivity: $$f(x,y) = f(a,b)\Rightarrow \dfrac{x^2+y^2}{(|x|+|y|)^2} = \dfrac{a^2+b^2}{(|a|+|b|)^2}.$$
But $x^2+y^2=a^2+b^2=1$, so $|xy| = |ab|$ or $x^2y^2 = a^2b^2\Rightarrow a^2+b^2-x^2-\dfrac{a^2b^2}{x^2}=0\Rightarrow (x^2-a^2)(x^2-b^2)=0,$ when $x\neq 0.$ If $|x|=|a|$, then $|y| = |b|$ and $\dfrac{x}{|x|+|y|} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Show that $g(x) = \log_2x$ for $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$
Show that $g(x) = \log_2x$ for $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$
In other words, simplify $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$ into $\log_2x$
I did:
$$\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12) \\
= \log... | Hint:
$$12x+6x^2=6x(x+2)$$
$$6x+12=6(x+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Elementary Algebra The exercise is to solve this equation: $$ \frac{2}{x} + \frac{3}{y} = 1,\quad x, y \in \mathbb{Z}^{*}$$
I tried as follows:
$$ \frac{2}{x} = 1 - \frac{3}{y} \implies \frac{2}{x} = \frac{y-3}{y} \implies x = \frac{2y}{y-3} \in \mathbb{Z}^{*}, \quad y \neq 3 $$
$$ \frac{3}{y} = 1 - \frac{2}{x} \imp... | Note that if $x,y\in \mathbb{Z}$ and $x\not=0$, $x\not=2$,
$$y = \frac{3x}{x-2} =3+\frac{6}{x-2}\Leftrightarrow \mbox{ $(x-2)$ divides 6}\Leftrightarrow (x-2)\in\{\pm 1, 2,\pm 3,\pm 6\}\\
\Leftrightarrow x\in\{-4,-1,1,3,4,5,8\}.$$
Hence the equation has $7$ solutions:
$$(-4, 2),\;
(-1, 1),\;
(1, -3),\;
(3, 9),\;
(4, 6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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proof "$cos^2(\frac{x}{2})=\frac{1+cos(x)}{2}$" I want to proof:
$cos^2(\frac{x}{2})=\frac{1+cos(x)}{2}$
I have changed the given eqation $1 = cos^2(x)+sin^2(x)$
$\to$ $cos^2(x) = 1- sin^2(x)$
Then another given eqation:
$cos(2x) = cos^2(x)-sin^2(x)$ $\to$ $-sin^2(x)=cos(2x)-cos^2(x)$
After that I have put those equt... | Usually the easiest way to proof trigonometric identities is to use complex exponentials. Here we have $\cos(x) = \frac{1}{2}(e^{ix}+e^{-ix})$ and thus:
$$ \cos^2(\frac{x}{2})
= \frac{1}{4}(e^{i\frac{x}{2}}+e^{-i\frac{x}{2}})^2
= \frac{1}{4}(e^{ix} + 2+ e^{-ix}) = \frac{1}{2}+\frac{1}{2}\cos(x) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Finding the Units Digit to 7 to the 2945 How would I go about finding the Units Digit to 7^(2945)?
I know that:
7^0 = 1
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
...
7^9 = 40353607
| when u are interested only in the unit digit you can multiply step by step like u did in your post but you can forget the other digits.
So you have $$7^2 = 49$$ however you forget what was it exactly and u only remember that $$7^2 = x\cdot 10 + 9 $$
you multiply again
$$7^3=7^2 \cdot 7 = 70x+63 = (7x+6) \cdot 10+3$$
ye... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How do I prove this using mathematical induction? \begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n(n+2) = \frac{n(n+1)(2n+7)}{6}
\end{align}
Using the mathematical induction step I arrive at this :
\begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n+1(n+3) = \frac{n+1(n+2)(2n+9)}{6}
\end{align}
And I don't see any other way to cont... | You have missed some parenthesis.
Write the induction step as:
$$
1\cdot 3 +2\cdot 4+\cdots +n(n+2)+(n+1)(n+3)=
$$
$$
=\frac{n(n+1)(2n+7)}{6}+(n+1)(n+3)=
$$
$$
=\frac{n(n+1)(2n+7)+6(n+1)(n+3)}{6}=\frac{(n+1)(2n^2+13n+18)}{6}=
$$
$$
=\frac{(n+1)(2n^2+4n+9n+18)}{6}=\frac{(n+1)[(2n(n+2)+9(n+2)]}{6}=
$$
$$
=\frac{(n+1)(n+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Complex solutions of $z^3 + 8 = 0$ (help with resolution method) How do I solve the equation $z^3 + 8 = 0 *($z$ being a complex number)* using polar coordinates?
These are the first lines of a solution method I saw:
$$z^3 + 8 = 0$$
$$\Longleftrightarrow z^3 =-8=8e^{i\pi+ 2 \pi in}$$ ($n$ is a natural number)
Could you ... | $$
z^3 = -8
$$
In polar coordinates you have $z = r e^{i\theta} = r(\cos\theta+i\sin\theta),$ where $\theta$ is real and $r$ is real and nonnegative.
A basic fact you need here is that $z^n = r^n(\cos(n\theta) + i\sin(n\theta)).$ If $n=3$ that says $z^3 = r^3(\cos(3\theta)+i\sin(3\theta)).$
Thus $-8 = 8\cdot( -1 + (i\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the Sum of an Infinite Series Compute
$${1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.}$$
I tried creating a partition for this but no such luck.
What would the the equation to generate the sum for the nth term
| One may recall the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ multiplying by $x$ and making $n \to +\infty$ in $(2)$, using $|x|<1$, g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
$\text{lcm}(a,b)+7\gcd (a,b) = a^2+b^2$
Find all positive integers such that $\text{lcm}(a,b)+7\gcd (a,b) = a^2+b^2$.
I thought about setting $a = ck$ and $b = dk$, where $\gcd(c,d) = 1$, which makes our expression become $$cdk+7k = (ck)^2+(dk)^2.$$ Equivalently, $$cd+7 = k(c^2+d^2).$$ How do we continue from here or... | We must find all pairs of coprime positive $c,d$ such that $c^2+d^2| cd+7$.
( we can let $c\leq d$ to ease calculations)
Notice that $c^2+d^2\geq 2cd$, so we must have $cd\leq 7$.
The only pairs of coprime positive integers $c,d$ with $cd\leq 7$ are:
$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(2,3)$
Of course $1+d^2> d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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How do you solve the equation $\frac{1}{4}(8^{2x+1})+5\cdot8^x - 3 = 0$? $$
\frac{1}{4}(8^{2x+1})+5\cdot8^x - 3 = 0
$$
My work:
$t = 8^x$
$\frac{1}{4}(t^{2+1})+5t - 3 = 0$
Then I don’t know how to do it, but I think I did it wrong.
| $\frac{1}{4}(t^2+1) +5t - 3 = 0$
$(t^2+1) + 20t - 12= 0$
$ t^2 + 20t - 11= 0$
Find it's roots using suitable method.
Here are some methods
And for this question you have to use quadratic formula.
Edit :
You have two different equation in your question. According to other solution is -
$\frac{1}{4}(8^{2x}\cdot8) +5\cdot... | {
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"url": "https://math.stackexchange.com/questions/2060248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Mathematical Induction and Divisibility $7\mid 11^n-4^n$ Prove $11^{n} - 4^{n}$ is divisible by 7 for all natural numbers n.
I know this holds for the base case of n = 1. Am I supposed to use $11^{n + 1} - 4^{n + 1}$ for the induction step?
Any help would be greatly appreciated!
Thank you in advance!
| Another way to to do it without induction is to note that the mod cycles for powers of $11$ and $4$ modulo congruent to $7$ correspond to each other.
$$\begin{equation}\begin{aligned}11 &\equiv 4 \pmod 7\\ 11^2 &\equiv 2 \pmod 7\\ 11^3 &\equiv 1 \pmod 7 \\ 11^4 &\equiv 4 \pmod 7 \\ &\vdots\end{aligned}\end{equation}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Prove that $\sum\limits_{cyc}\frac{a^2}{a^3+2}\leq\frac{4}{3}$ if $a, b, c, d > 0$ and $abcd=1$
Let $a$, $b$, $c$ and $d$ be positive numbers such that $abcd=1$. Prove that:
$$\frac{a^2}{a^3+2}+\frac{b^2}{b^3+2}+\frac{c^2}{c^3+2}+\frac{d^2}{d^3+2}\leq\frac{4}{3}.$$
Vasc's LCF Theorem does not help here. Also I tried ... | Here is a solution that relies on an uninspiring chain of single-variable inequalities of the form $\frac{x^2}{x^3+2}\leq \alpha+\beta\ln(x)$ on some interval for various nonnegative real $\alpha$ and $\beta$.
Define $f(x)=\frac{x^2}{x^3+2}$. Note first that $f(x)$ is maximized when
$$f'(x)=\frac{2x(x^3+2)-3x^2(x^2)}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Prove $|\cos z|^2=\cos^2x+\cosh^2y$ That what I tried
$$
\cos z= \cos (x+iy)= \cos x \cos iy- \sin x \sin iy=\cos x \cosh y-i\sin x \sinh y\\
|\cos z|^2= \cos^2(x) \cosh^2(y)+\sin^2(x)\sinh^2(y)\\
\sinh^2(y)=\cosh^2(y)-1
$$
Then
$$
\cos^2(x) \cosh^2(y)+\sin^2(x) \cosh^2 (y)-\sin^2 (x)= \cosh^2 (y)-\sin^2 (x)
$$
What ... | The formula you want to prove is wrong: for $z=0$, we have $\lvert\cos0\rvert^2=1$, whereas $\cos^20+\cosh^20=2$.
\begin{align}
\lvert\cos z\rvert&=
\cos^2x\cosh^2y+\sin^2x\sinh^2y\\
&=\cos^2x\sinh^2y+\cos^2x+\sin^2x\sinh^2y\\
&=\cos^2x+\color{red}{\sinh^2y}
\end{align}
Your formula is correct as well.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the smallest n for inequality involving factorial, $n! > 10^6 $, what is $n$? How to find the smallest n for inequality involving factorial, $n! > 10^6 $, what is n?
I solved it using calculator, that is the answer is $n \approx 9$. How to solve this inequality?
Smallest $n$ giving $n! > 10^6$ ?
Edit
... | $$
\begin{align*}
9!&=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\\
&=(2\cdot5)(3\cdot7)(4\cdot6)(8\cdot9)\\
&<10\cdot 30\cdot 30\cdot 80\\
&<8\cdot10^5\\
&<10^6.
\end{align*}
$$
and
$$
\begin{align*}
10!&=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\\
&=(2\cdot5)(3\cdot7)(4\cdot6)(8\cdot9)\cdot10\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Work and time problem . People leaving work one by one and at the same time reducing their rate of doing work.
Question: A group of $30$ people can complete a job working for $10$ hours a day in
$15$ days. The group starts a work. But at the end of every day, starting
from the first day, one person leaves the grou... | Needed amount of work:
$$
W
= 30 \cdot 10 \frac{\text{h}}{\text{d}} \cdot 15 \text{d}
= 4500 \text{h}
$$
Rate of work per person over time:
$$
r(t) = 10 \frac{\text{h}}{\text{d}}
$$
Active people:
\begin{align}
n(1) &= 30 \\
n(t) &= 30 - (t - 1) = 31 - t
\end{align}
Work done until day $t$, where the group in total ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limits - a different approach $\lim_{x \to\infty }(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$.
How do I calculate the value of l and m?
My try: I know questions having limit tending to infinity can be solved by dividing the numerator and denominator by greatest power of $x$.But it got me nowhere in this question. Any he... | $$\lim\limits_{x \to \infty}(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$$
$$\lim\limits_{x \to \infty}(\frac{x^3+4x^2+3x+5+lx^3+2lx^2+3lx+mx^2+2mx+3m}{x^2+2x+3})=10$$
$$\lim\limits_{x \to \infty}(\frac{(1+l)x^3+(4+2l+m)x^2+(3+l+2m)x+(5+3m)}{x^2+2x+3})=10$$
Note that as the limit exists, the coefficient of $x^3$ has to be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Finding out the limit: $\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n}\sqrt {k(n-k)}$ Find out the limit of the sum $\lim \frac{1}{n}\sum_{k=0}^{n}\sqrt {k(n-k)}$ as $n$ tends to $\infty.$
| Since $x-x^2$ has a peak at $x=\frac{1}{2}$ and decreases monotonically to $0$ as $x \downarrow 0$ or $x \uparrow 1$, we see $\frac{1}{n}\sum_{k=0}^n \sqrt{k(n-k)} = \sum_{k=1}^{n-1} \sqrt{\frac{k}{n}(1-\frac{k}{n})} \ge \sum_{k=1}^{n-1} \sqrt{\frac{1}{n}(1-\frac{1}{n})} = (n-1)\sqrt{\frac{1}{n}-\frac{1}{n^2}} = \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Factoring a polynomial of degree 5 I was looking at the solutions to a question about BIBO stability. The transfer function was
$$\dfrac{s^2+1}{s^5+2s^4+4s^3+7s^2+3s+5}=\dfrac{1}{s^3+2s^2+3s+5}$$
So the lecturer factorised the bottom polynomial and cancelled accordingly, but it isn't explained in the solutions. Is the... | There exists as simple method to find quadratic factors of monic polynomials of any degree that doesn't involve anything more than straightforward high school level algebra. For 5th degrees polynomials this works as follows. We want to factorize the function:
$$f(x) = x^5 + a x^4 + b x^3 + c x^2 + d x + e$$
where the c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
finding value of indefinite integration of $\frac{y^7-y^5+y^3-y}{y^{10}+1}dy$ finding value of indefinite integration $\displaystyle \int\frac{y^7-y^5+y^3-y}{y^{10}+1}dy$
$\displaystyle \int\frac{y^5(y^2-1)+y(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2+1-2)}{y^{10}+1}dy$
$\displ... | hint: put $u = y^2$, and note that $(y^5 + y)dy = y(y^4+1)dy = \dfrac{u^2+1}{2}du$. Would this enable you to continue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Functional equation $(x + y)f(f(x)y) = x^2 f(f(x) + f(y))$ Find all functions $f:\mathbb{R}^+ \to \mathbb{R}^+$ such that
$$(x + y)f(f(x)y) = x^2 f(f(x) + f(y))
\mbox{, for all } x,y\in\mathbb{R}^+.$$
I tried out various substitutions such as $x=y$, $x=1$, $x+y=x^2$, and nothin' works.
Thanks in advance for your reply.... | \enspace \frac{f(af(b))}{a}=\frac{f(af(a))}{a} = (3) = \frac{f(2f(a))}{2} = \frac{f(2f(b))}{2} = (3) = \frac{f(bf(b))}{b}=\frac{f(bf(a))}{b}\\
B: \enspace (2) \implies \frac{f(af(b))}{b^2} = \frac{f(bf(a))}{a^2} \\
A \enspace and \enspace B: \enspace \implies \frac{a}{b} \cdot \frac{f(bf(a))}{b^2} = \frac{f(af(b))}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Simplify $\frac{1}{\sqrt{-1+\sqrt{2}}}$ In an exercise I got as solution $\frac{1}{\sqrt{-1+\sqrt{2}}}$, it now holds that $\frac{1}{\sqrt{-1+\sqrt{2}}} = \sqrt{1+\sqrt{2}}$. But I really don't see how you can manipulate the left hand side to become the right hand side (quite shameful as a mathematician, I must admit).... | You are given the fraction: $$\frac{1}{\sqrt{-1 + \sqrt{2}}} \tag1$$
Square it: $$\bigg(\frac{1}{\sqrt{-1 + \sqrt{2}}}\bigg)^2 = \frac{1^2}{\sqrt{-1 + \sqrt{2}}^2} = \frac{1}{-1 + \sqrt{2}} \tag2$$
Multiply by $1$: $$\frac{1}{-1 + \sqrt{2}}\times 1 = \frac{1}{-1 + \sqrt{2}}\times \frac{-1 - \sqrt{2}}{-1 - \sqrt{2}} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve system of equations with square roots and fractions $$
\begin{cases}
\left(\frac{1+\sqrt{3}}{2}\right)^2x+\left(\frac{1+\sqrt{3}}{2}\right)y+1=0\\
\left(\frac{1-\sqrt{3}}{2}\right)^2x+\left(\frac{1-\sqrt{3}}{2}\right)y+1=0
\end{cases}
$$
I tried setting the equations equal to each other and substituting variables... | swetting $$a=\frac{1+\sqrt{3}}{2}$$ and $$b=\frac{1-\sqrt{3}}{2}$$ then we have to solve
$$a^2x+ay=-1$$
$$b^2x+by=-1$$
from the first equation we get
$$y=-\frac{1+a^2x}{a}$$
plugging this in the second equation we get
$$a^2bx-b(1+a^2x)=-1$$ thus
$$x=\frac{b-1}{ab^2-a^2b}$$
can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Power Series involving Double Factorials This comes from my partial solution to another question. I need to find a closed form for the following summation
$$\sum_{n=1}^{\infty}\frac{n!}{(2n+1)!!}n^x$$
Where $x$ is a fixed integer
This appears similar to many arcsine series and formulas
Particular values:
$$\begin{... | Why just even powers of $n$? Let
$$ F(j) = \sum_{n=1}^\infty \frac{n! n^j}{(2n+1)!!} = \sum_{n=1}^\infty \frac{2^{-1-n} \sqrt{\pi} n! n^j}{\Gamma(n+3/2)} $$
The exponential generating function is
$$\eqalign{E(z) &= \sum_{j=0}^\infty \frac{F(j)}{j!} z^j\cr
&= \sum_{n=1}^\infty \frac{2^{-1-n} \sqrt{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Computing $16^{74} \bmod 65$ I have the following problem. I need to calculate $16^{74} \bmod 65$. Is there a way to do this without a calculator?
Thank you for your help in advance!
EDIT (M.S.): Notice that this question is linked as an example in one of the answers to the FAQ post: How do I compute $a^b\,\bmod c$ by ... | Repeatedly raising to as small powers as possible and immediately reducing modulo $65$ we have
$$(16)^{74} = (16^2)^{37} = (256)^{37} = (260 - 4)^{37} \equiv (-4)^{37} = - 4^{37} = -4 \cdot 4^{36} = -4 \cdot (4^4)^9 = \\
-4 \cdot (256)^9 = -4 \cdot (260 - 4)^9 \equiv -4 \cdot (-4)^9 = 4^{10} = 4 \cdot (4^3)^3 = 4 \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 1
} |
Square root positive/negative What is the square root of $9$? Is it always $\pm 3$ or just positive $3$?
Trying to find the solution set of this equation : $x-3 = \sqrt{x+3}$
I want to understand the concept of square root to solve the problem.
Thanks
| There are two numbers $x$ such that $x^2=9$ : $x=3$ and $x=-3$, but with the symbol $\sqrt{9}$ we indicate only the positive one, so $\sqrt{9}=3$.
In other words, the symbol $\sqrt{x}$ indicates, by definition, the positive number $y$ such that $y^2=x$.
This implies that if we search the solution of the equation $x-3=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Similar triangles: the shortest side of triangle $B$ measures $6$. Find the other two sides. Two triangles are similar. Triangle $A$ has sides measuring $3$, $4$, and $5$. The shortest side on triangle $B$ measures $6$. What are the lengths of the other two sides of triangle $B$?
| Let $x,y$ be respectively the lengths of the medium and longest sides of triangle "$B$". Then similarity of triangles implies: $\dfrac{3}{6} = \dfrac{4}{x} = \dfrac{5}{y}\implies x = 8, y = 10.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int \frac{dx}{\sin^2 (x) + \sin(2x)}$ We have to do the following integration
$$\int \frac{dx}{\sin^2 (x) + \sin(2x)}$$
I tried as
In the numerator, $1 = \sin^2(x)+\sin(2x)+\cos^2(x)-\sin(2x)$
But now how to proceed?
| \begin{align}
\int\frac{dx}{\sin^2(x)+\sin(2x)}&=\int\frac{\sec^2(x)}{\sin^2(x)\sec^2(x)+2\sin(x)\cos(x)\sec^2(x)}dx\\
&=\int\frac{\sec^2(x)}{\tan^2(x)+2\tan(x)}dx
\end{align}
Letting $u:=\tan(x)$, then $du=\sec^2(x)dx$ gives
\begin{align}
\int\frac{dx}{\sin^2(x)+\sin(2x)}&=\int\frac{du}{u^2+2u}\\
&=\int\frac{du}{(u+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Is this a subspace with the given eigenvectors as its basis vectors? Given the eigenvectors:
$$\begin{pmatrix} 2 \\ 3 \end{pmatrix}, \begin{pmatrix}4 \\ 5\end{pmatrix}$$
And this linear combination equation
$$ \begin{pmatrix} 2 \\ 4 \end{pmatrix} = c_1 \begin{pmatrix} 2 \\ 3 \end{pmatrix}+c_2\begin{pmatrix}4 \\ 5\end{p... | Note that $\frac{1}{2} \left[ 3 \begin{pmatrix} 4 \\ 5 \end{pmatrix} - 5 \begin{pmatrix} 2 \\ 3 \end{pmatrix} \right] = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, so the space with basis vectors $\Bigg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove $\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx={1\over 2}$ using an alternative method Prove that
$$\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx={1\over 2}\tag1$$
My try:
$u=1+\sqrt{5}x$ then $du=\sqrt{5} \, dx$
$${1\over \sqrt 5}\int_1^{1+\sqrt{5}}(4u^{-3}-u^{-2}) \, du$$
$$\left. {1\over \sqrt{5}}(-2u^{... | Set $\frac{3-\sqrt{5}x}{1+\sqrt{5}x}=u$, we have $\frac{-4\sqrt{5}}
{(1+\sqrt{5}x)^2} dx=du$ thus
$$\int_{0}^{1} {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx=\frac{1}{-4\sqrt{5}}\int_{3}^{\sqrt{5}-2}u\,du=\frac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Deriving a Series Representation of an Integral (Riemann Zeta Function) It is known that
$$\int_0^\infty\frac{x^n}{e^x-1}dx=n!\zeta(n+1)$$
for integer values of $n$ (this is also generalises, but that's not important for this question). I have also had a look at how to arrive at this expression, starting from the serie... | Your so called proof is very much unorthodox. A much shorter proof can be found in this post and it basically goes as follows:
$$\begin{align}\int_0^\infty\frac{x^k}{e^x-1}\ dx&=\int_0^\infty\frac{x^ke^{-x}}{1-e^{-x}}\ dx\\&=\int_0^\infty x^ke^{-x}\sum_{n=0}^\infty e^{-nx}\ dx\\&=\int_0^\infty\sum_{n=1}^\infty x^ke^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If, in a triangle, $\cos(A) + \cos(B) + 2\cos(C) = 2$ prove that the sides of the triangle are in AP By using the formula :
$$
\cos(A)+\cos(B)+\cos(C) = 1 + 4 \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)
$$
I've managed to simplify it to :
$$
2\sin\left(\frac{A}{2}\right)\sin\... | Hint:
Prosthaphaeresis Formulas: $\cos A+\cos B=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2$
Double angle formula: $1-\cos C=2\sin^2\dfrac C2$
$\dfrac{A+B}2=\dfrac\pi2-\dfrac C2\implies\cos\dfrac{A+B}2=\sin\dfrac C2$
As $0<c<\pi, \sin\dfrac C2\ne0$
So, we have $\cos\dfrac{A-B}2=2\cos\dfrac{A+B}2$
Expand $\cos\dfrac{A\pm B}2$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Help in understanding a simple proof. Let $Ax + By + C = 0$ be a general equation of a line and $x\cos \alpha + y\sin \alpha - p = 0$ be the normal form of the equation.
Then,
$${-p\over C } = { \cos \alpha\over A} = { \sin\alpha\over B}\tag{1}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} \tag{2}$$... | $$
{p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B}
\\
\implies \frac{p^2}{C^2}=\frac{\cos^2(\alpha)}{A^2}=\frac{\sin^2(\alpha)}{B^2}
$$
If
$$
\frac{a}{b}=\frac{c}{d}=k
\\
\implies a=bk, c=dk
$$
So the value of :
$$
\frac{a+c}{b+d}=\frac{bk+dk}{b+d}=\frac{k(b+d)}{b+d}=k=\frac{a}{b}=\frac{c}{d}
$$
Simply p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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A combinatorial identity with binomial coefficients and floor function. Show that $$ \sum_{m=0}^{2k+1}{2^{m}\binom{n}{m}\binom{n-m}{\bigl\lfloor \frac{2k+1-m}{2} \bigr\rfloor}}=\binom{2n+1}{2k+1} $$
I tried expending the sum and using induction, but could not complete the induction step; I tried using Pascal's identity... | Suppose we seek to prove that
$$\sum_{k=0}^{2m+1} {n\choose k} 2^k
{n-k\choose \lfloor (2m+1-k)/2 \rfloor}
= {2n+1\choose 2m+1}.$$
Observe that from first principles we have that
$${n\choose \lfloor q/2 \rfloor} = {n\choose n-\lfloor q/2 \rfloor} =
\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}}
\\ \times \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Dividing rational expression? Just to clarify, I do not want anyone to do the problem for me. I just need someone to explain this stuff to me because I'm completely lost, and I would really appreciate it if you could take the time to do so.
In the book I've been reading they have this example:
$$\frac{y^3-1}{y^2-1} \d... | $(y^3-1)/(y^2-1) ÷ (y^2 + y + 1)/(y^2 + 2y + 1)=$
$(y^3-1)/(y^2-1)\times (y^2 +2 y + 1)/(y^2 + y + 1)=$
$\frac {(y^3 -1)(y^2+2y+1)}{(y^2-1)(y^2+y+1)}=M$
I think you've followed and figured all that on your own.
Now comes.... experience, intuition and elbow grease.
Does $y^3 - 1$ factor? I know it does because I've bee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Integral $\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ We have to evaluate the following integral:
$$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$
I tried this:
I multiplied both the numerator and denominator by $\sec x$
And substituted $\tan x = t$.
But after that I got stuck.
The book ... | $$\displaystyle\int\frac{\sqrt{\sin x}dx}{\sqrt{\sin x}+\sqrt{\cos x}}\;=\;\int\frac{dx}{1+\sqrt{\cot x}},\\
\displaystyle\cot x=z^{2} \;\;\Rightarrow\;\; -(1+z^{2})dx=2zdz,\\
\displaystyle\int\frac{dx}{1+\sqrt{\cot x}} \;=\; -\int\frac{2zdz}{(1+z^{2})(1+z)} \;=\; -2\int\frac{z(z-1)dz}{z^{4}-1}.\\
\displaystyle\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
How is $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \sqrt{2}-1$? I wondered, why this:
$$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$$
is equal to $\sqrt{2}-1$.
Can anyone explain me, why this is equal? :/
| \begin{align}
\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} &= \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\cdot\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}-1}}\\
&=\sqrt{\frac{\left(\sqrt{2}-1\right)^2}{2-1}}\\
&=\sqrt{\left(\sqrt{2}-1\right)^2}\\\\
&=\sqrt{2}-1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Probability of selecting a ball from a box. Suppose we have $12$ identical balls. Think of the balls as having distinct ID numbers, $1-12$. We have $3$ identical boxes, each of which can contain exactly $4$ balls. One after the other the balls are thrown at random into one of the three boxes.
Suppose Mr. A will buy one... | For $1\lt j\lt k$, there are $3!\binom{12-k}{3}\binom{8-j}{3}$ ways to buy $1,j,k$.
$3!$ due to the arrangements of $1,j,k$. $\binom{12-k}{3}$ due to the number of ways to choose $3$ balls greater than $k$. $\binom{8-j}{3}$ due to the number of ways to choose $3$ balls greater than $j$ from those left over from choosin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$
What is $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ ?
Here are a few remarks:
*
*Since $x\mapsto \frac{2^x}{x}$ is increasing when $x\geq 2$, one might be tempted to use the integral test. This fails: when doing so, one gets $a_n\leq \sum_{k=1}^n \fra... | Let
$$ S_n=\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}. \tag{1}$$
Clearly
$$ S_n=\sum_{k=1}^n\frac{n}{k}\frac1{2^{n-k}}\ge \sum_{k=1}^n\frac1{2^{n-k}}=2-\frac{1}{2^n}. $$
Define
\begin{eqnarray}
f_n(x)=\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}\left(\frac12x\right)^k
\end{eqnarray}
and then $f_n(0)=0,f_n(2)=S_n$ and
$$ f_n'(x)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Prove that $\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$ Prove that:
$$\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$$
My Work:
I guess that I have to use the formula :
$$\tan A = \frac {2 \tan(\frac {A}{2})}{1-\tan^2 (\frac {A}{2})}$$
But, I am not being able to use it. Please help me.
| Hint: We have $$\tan 7.5^\circ =\frac {\sin 7.5^\circ}{\cos 7.5^\circ} =\frac {2\sin^2 7.5^\circ}{2\cos 7.5^\circ \sin 7.5^\circ} =\frac {1-\cos 15^\circ}{\sin 15^\circ} =\frac {1- \cos (45^\circ -30^\circ)}{\sin (45^\circ -30^\circ)} $$
Can you take it from here? Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Limit of multivariable $\frac{x^2+y^2}{ x^4+y^4}$ Find the limit
$$\lim_{(x, y) \to (0,0)} \frac {x^2+y^2 }{x^4+y^4 }$$
This limit does not exists since when convert it into polar we get $\frac {1 }{ r^2 (1-2 \sin^2 \theta \cos^2 \theta)}$ which is one over zero right?
| plugging $$x=\frac{1}{\sqrt{n}},y=\frac{1}{n^2}$$ in the term above we get
$${\frac { \left( n+1 \right) \left( {n}^{2}-n+1 \right) {n}^{4}}{
\left( {n}^{2}+1 \right) \left( {n}^{4}-{n}^{2}+1 \right) }}
$$ and the limit is $$+\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Points $E$, $F$ and $G$ lies on median of triangle $ABC$. Determine the ratio $|GF|:|EF|$ In a given triangle $ABC$ lets choose inside side $AC$ points $P$, $Q$ and inside side $BC$ points $R$,$S$ so, that $|AP|=|PQ|=|QC|$, $|BR|=|RS|=|SC|$
Next denote point $E$ as intersection of diagonals in trapezoid $ABRP$, point $... | First off, let's label the intersections of the median with $QS$, $PR$, and $AB$ as $I$, $H$, and $K$, respectively. Note that by the three parallels theorem, $CI = IH = HK$, so let's set the length of $HK = x$.
Note also that triangles $CQS$, $CPR$, and $CAB$ are similar, and using the appropriate ratio of similariti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The quadratic equation $x^2+x=3kx-k^2$ has two different real roots. Find the range of $k$
The quadratic equation $x^2+x=3kx-k^2$ has two different real roots. Find the range of $k$.
My answer is $k<1$ or $k<\frac{1}{5}$, but the answer sheet says $k<\frac{1}{5}$ or $k>1$.
What have I done wrong? Please help.
What I'... | The discriminant of $$x^2+x(1-3k)+k^2=0$$
$$(1-3k)^2-4k^2=1-6k+5k^2=5\left(k-\dfrac15\right)(k-1)$$ which needs to be $\ge0$
Now for $(y-a)(y-b)\ge0$ with $a\le b$
If $y-a\ge0\iff y\ge a;$ we need $y-b\ge0\iff y\ge b\implies y\ge$ max$(a,b)=b$
What if $y-a\le0?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluating an expression given values of symmetric polynomials Evaluate $\dfrac x{yz} + \dfrac y {xz} + \dfrac z y$
Given, $z+y+x=4, \qquad xyz=-60, \qquad xy+xz+yz=-17$
How do we do this? I found a common denominator, and substituted it for $-60$, but I am unaware of how to proceed.
Someone already asked the question... | Assuming there is a typo.
We have to find -
\begin{align}
& \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \\[10pt]
= {} & \frac{x^2 + y^2 + z^2}{xyz} \tag 1
\end{align}
Also $x + y + z = 4$.
Squaring both sides,
$$x^2 + y^2 + z^2 + 2(xy + yz + zx) = 16$$
$$x^2 + y^2 + z^2 + 2(-17) = 16$$
$$x^2 + y^2 + z^2 = 16 + 34$$
$$x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Maximum of $\frac{1+3a^2}{(a^2+1)^2}$ What is the maximum value of
$\displaystyle{{1 + 3a^{2} \over \left(a^{2} + 1\right)^{2}}}$, given that $a$ is a real number, and for what values of $a$ does it occur ?.
| Let $x=a^2\ge0$. We then have to look at
$$\frac{1+3x}{(x+1)^2}$$
upon differentiating and setting equal to $0$, we get
$$\frac{3(x+1)^2-2(x+1)(1+3x)}{(x+1)^4}\\\implies3(x+1)^2=2(x+1)(1+3x)\\\implies3(x+1)=2(1+3x)\\\implies x=\frac13$$
Thus, the relative maxima (or minima) occurs at
$$a=\pm\sqrt x=\pm\frac{\sqrt3}3$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How was this equation derived?
Does anyone know how the equation in red was derived?
| First part:
$$\cos\theta=\cos^2\frac\theta2-\sin^2\frac\theta2=\left(\cos^2\frac\theta4-\sin^2\frac\theta4\right)^2-4\sin^2\frac\theta4\cos^2\frac\theta4.$$
Second part:
$$\cos^2\frac\theta4=\frac1{1+\tan^2\frac\theta4}=\frac1{1+t^2},\\
\sin^2\frac\theta4=\tan^2\frac\theta4\cos^2\frac\theta4=\frac{t^2}{1+t^2}.$$
But $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Factorizing $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ The intermediate factorization calculation $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ was coined straightforward by Riley's Mathematical Methods for Physics and Engeneering (eq.24.66).
I've tried completing the square,
$$
z^2-z\frac{a^2 +b^2}{ab}+1 = 0 \\
\b... | Note that
$$\frac{a^2+b^2}{ab}=\frac ab+\frac ba$$
and
$$\frac ab\cdot\frac ba=1$$
Remember that, in a polynomial of second degree, the coefficient of $z$ is the opposite of sum of the roots and the costant term is their product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solutions of $ 2- \frac{1}{z} = z^* $ in $\mathbb{C}$ I am asked to show that the only solution of the equation $ 2- \frac{1}{z} = z^* $ is $z=1$.
I have tried this a few ways the two where i get closest are as follows.
$\frac{2z-1}{z}=z^*$
$0=z^* z -2z +1$
$\left|z^2 \right| -2z +1=0 = \left|z \right|^2 -2z +1 $
id r... | $$\left|z^2 \right| -2z +1=0 \\ \to \sqrt{x^2+y^2} ^2 -2(x+iy) +1=0\\
x^2+y^2-2x+1-i(2y)=0 $$ I t is better to say $$x^2+y^2-2x+1-i(2y)=0 +i0 \to \\
\begin{cases}x^2+y^2-2x+1=0 & (x-1)^2+y^2=0\\-2y=0 & \to y=0 \end{cases}\\
\begin{cases}x^2+y^2-2x+1=0 & (x-1)^2+0^2=0 \to\\-2y=0 & \to y=0 \end{cases}\\(x-1)^2=0 \to x=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solutions to $|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|$ Find all the complex numbers $z$ satisfying
\begin{equation}
|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|
\end{equation}
I tried using $z=a+bi$, then using the formula for absolute value:
\begin{equation}
|(a+1)+i(b-\sqrt{3})|=|(a-1)+i(b+\sqrt{3})|\\
\sqrt{(a+1)^2+(b-\sqrt{3})^2}=\sqrt... | Yes, this is a correct way to approach the problem. By squaring each side of your last line and simplifying, one obtains:
$$ 4a - 4\sqrt{3}b = 0 \Rightarrow a = \sqrt{3} b.$$
Thus the set that you are looking for consists of all the complex numbers $z = a + bi$ such that $a = \sqrt{3}b$, which in the complex plane is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
A binomial identity from counting permutations with cycles of length two. Let $n\ge 1$ be an integer. Consider the following sequence:
where $\left\{l_j\right\}_{j=1}^n$ are indices.By analyzing all possible decompositions of the sequence above into distinct pairs we have discovered the following identity:
\begin{equa... | Suppose we seek to prove that
$$n! \sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{(2q-1)!!}{2^q q!} {n\choose 2q} = (2n-1)!!,$$
which is
$$n! \sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{1}{2^q q!} \frac{(2q)!}{2^q q!}
{n\choose 2q} = \frac{(2n)!}{2^n n!}$$
or
$$\sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{1}{4^q} {2q\choose q} {n\choose ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Basel's problem: $5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}=\zeta(2)$ Golden ratio:$\phi$
$$5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}=\zeta(2)\tag1$$
We'll known Basel's problem
$$\sum_{k=1}^{\infty}{1\over k^2}=\zeta(2)\tag2$$
How do I transform $(1)$ to $(2)?$
Extra informati... | You want to show that
$5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}
=\zeta(2)\tag1
$.
The function you need
is the dilogarithm,
defined by
$L_2(z)
=\sum_{k=1}^{\infty}\dfrac{z^k}{k^2}
$.
There is much useful info here:
https://en.wikipedia.org/wiki/Spence's_function
So you want to show that
$5(L_2(\ph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Check Convergence of the series: $\sum_{n=1}^\infty {\frac{ \sqrt{n+1} - \sqrt{n}}{\sqrt{n}}}$ I have to check the convergence of this series:
$$\sum_{n=1}^\infty {\frac{ \sqrt{n+1} - \sqrt{n}}{\sqrt{n}}}$$
Which is equal to $$\sum_{n=1}^\infty \frac{\sqrt{n+1}}{\sqrt{n}} - 1$$
What can I do here to check whether this ... | Write the partial sum as
$$s_m = \sum_{n=1}^{m}\left( \sqrt{1+\frac{1}{n}}-1\right)$$
For $n\ge 1$ we have the inequality
$$\left(\frac{1}{n}+1\right)-\left(-\frac{1}{8 n^2}+\frac{1}{2 n}+1\right)^2=\frac{8 n-1}{64 n^4}>0$$
whence by taking the square root
$$\sqrt{\frac{1}{n}+1}>-\frac{1}{8 n^2}+\frac{1}{2 n}+1\\
\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find all integers $(x, y)$ such that $1 + 2^x + 2^{2x + 1} = y^2$ Find all integers $$(x, y)$$ such that $$1 + 2^x + 2^{2x + 1} = y^2$$
So I basically used $$ f(x) = 1 + 2^x + 2^{2x + 1} = y^2$$ and created a table from 0 to 20. I got two pairs of integers: $$(0, \pm2)$$ and $$(4, \pm23)$$
I want to know if there's ano... | $1 + 2^x + 2^{2x + 1} = y^2 \\\implies 2^x(2^{x+1}+1) = (y+1)(y-1)$
$x<-1$ gives a non-integer LHS (no solutions)
$x=-1$ gives LHS $= 1$ with no solutions for $y$
$x=0$ gives LHS $= 3$ and $y=\pm 2$
For $x>0$, $y$ is odd so put $y=2k+1$ and $2^x(2^{x+1}+1) = (2k+2)(2k) = 4k(k+1)$ which is divisible by $8$ so $x\ge 3$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $z_1^2+z_2^2$ is real, $z_1(z_1^2-3z_2^2)=2$, and $z_2(3z_1^2-z_2^2)=11$, then find $(z_1^2+z_2^2)^2$
If $z_1$ and $z_2$ are complex numbers such that $z_1^2+z_2^2 \in\mathbb R$ and $$z_1(z_1^2-3z_2^2)=2,\qquad z_2(3z_1^2-z_2^2)=11,$$ then find the value of $(z_1^2+z_2^2)^2$. Given answer is $25$.
I have tried man... | I suggest that we square both equations.
$$z_1^2(z_1^2-3z_2^2)^2=4\\
z_2^2(3z_1^2-z_2^2)^2=121$$
Now add them together and simplify.
$$125=z_1^2(z_1^2-3z_2^2)^2+z_2^2(3z_1^2-z_2^2)^2=z_1^2(z_1^4-6z_1^2z_2^2+9z_2^4)+z_2^2(9z_1^4-6z_1^2z_2^2+z_2^4)=\\
=z_1^6+3z_1^4z_2^2+3z_1^2z_2^4+z_2^6=(z_1^2+z_2^2)^3$$
The rest is tri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\lim_{x\rightarrow a} (2-\frac{x}{a})^{\tan (\frac{\pi x}{2a})}$ Can someone help me to calculate the following limit?
$$\lim_{x\to a} \Big(2-\frac{x}{a}\Big)^{\tan\dfrac{\pi x}{2a}}$$
Thank you.
| Consider $\lim_{x\rightarrow a} e^{\ln((2-\frac{x}{a})^{\tan (\frac{\pi x}{2a})})}=\lim_{x\rightarrow a} e^{\tan(\frac{\pi x}{2a})\ln(2-\frac{x}{a})}$.
Clearly it suffices to calculate the limit $$\lim_{x\rightarrow a} {\tan(\frac{\pi x}{2a})\ln(2-\frac{x}{a})}.$$
We can do this by using L'Hopitals rule.
\begin{eqnarra... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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If $2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$ and $f^{-1}(x)=1$ then find the value of $x$ If $2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$ and $f^{-1}(x)=1$ then find the value of $x$.
My Attempt:
$$2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$$
$$2f(x)+3f(\frac {1}{x})=4x + \frac {6}{x}$$
At this point, I couldn't get othe... | You can solve the following system.
$2f(x)+3f\left(\frac{1}{x}\right)=\frac{4x^2+6}{x}$ and $2f\left(\frac{1}{x}\right)+3f(x)=\frac{\frac{4}{x^2}+6}{\frac{1}{x}}$,
which gives $f(x)=2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $x$ positive integer such that $x^4+x^3+x^2+x+1$ is a perfect square
Find the number of positive integers $x$ for which $x^4+x^3+x^2+x+1$ is a perfect square.
My attempts:
Let $x^4+x^3+x^2+x+1=k^2\implies (x+1)^2(x^2-x+1)=(k-x)(k+x)$
I analysed this a bit found $x=0$ as one which satisfy all condition, how do I ... | Note that the problem is equivalent to finding integer solutions to $$4y^2=4x^4+4x^3+4x^2+4x+4$$
Now proceed to note that if $x>3$, we can find $$(2x^2+x)^2=4x^4+4x^3+x^2 < 4x^4+4x^3+4x^2+4x+4=4y^2$$
And $$4y^2 < 4x^4+4x^3+5x^2+2x+1=(2x^2+x+1)^2 $$
Since $4x^4+4x^3+4x^2+4x+4$ is stuck between squares of two consecutive... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, the If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, theremainder is $x+a$, find $k$ and $a$.
My Attempt,
$f(x)=x^4-6x^3+16x^2-25x+10$
$g(x)=x^2-2x+k$
$R=x+a$
Here, the divisor is in the quadrati... | HINT
You can write:
$$x^4-6x^3+16x^2-25x+10= (x^2-2x+k)(x^2+bx+c)+x+a=\\
=x^4+(b-2)x^3+(c-2b+k)x^2+(-2c+bk+1)x+(a+kc)$$
So,
$$b-2=-6→b=-4\\
c-2b+k=16\\
-2c+bk+1=-25\\
a+kc=10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas? How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas?
I tried the $\pm \sqrt{\frac{1-\cos{A}}{1+\cos{A}}}$ and $\frac{\sin{A}}{1+\cos{A}}$ but got stuck once there were square roots on top and bottom like $\frac{\sqrt{...}}{1-\sqrt{...}}.$
Using th... | Fill in details:
$$r:=\tan\left(-\frac{5\pi}{16}\right)=\tan\frac{11\pi}{16}\implies$$
$$s:=\tan\frac{11\pi}8=\tan\left(2\cdot\frac{11\pi}{16}\right)=\frac{2\tan\frac{11\pi}{16}}{1-\tan^2\frac{11\pi}{16}}=\frac{2r}{1-r^2}$$
and
$$-1=\tan\frac{11\pi}4=\tan\left(2\cdot\frac{11\pi}8\right)=\frac{2s}{1-s^2}\implies$$
$$s^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Binoharmonic summation in sequence and series Please consider this bino-harmonic summation:-
nC0/a + nC0/a+d + nC0/ a+2d. . . . . . .nCn/a+nd
I tried a lot but could not get anything any way all of it was failure..Now I am beginning to think that I am a real dumbass ,please help me solve this..
| I believe you set out to compute
$$
\frac1a\binom{n}{0} - \frac{1}{a+d}\binom{n}{1} + \cdots + (-1)^n\frac{1}{a + nd}\binom{n}{n}.
$$
Consider the binomial expansion of $x^{a-1}(1-x^d)^n$:
$$
x^{a-1}(1-x^d)^n = x^{a-1}\left( \binom{n}{0} - \binom{n}{1}x^d + \cdots + (-1)^n\binom{n}{n}x^{nd} \right) \\
= \binom{n}{0}x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$ I'm trying to find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$. I think I'm mistaken and perhaps using the chain rule incorrectly.
Let $g(x) = \frac{2x}{1+x^2}$ and let $h(x) = \arcsin x$
According to the chain rule -
$$f'(x) = \frac{1}{\s... | Since it's easy to graph the function with some software (also online) it should be clear that the derivative cannot be everywhere positive and the graph also suggests the function is not differentiable at $-1$ and $1$.
Set $g(x)=\dfrac{2x}{1+x^2}$; then
$$
f(x)=\arcsin\dfrac{2x}{1+x^2}=\arcsin(g(x))
$$
and, by the ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to factor $x^3-1$ with mod $3$ when I factor $x^3-1$ with $\mod 3$ on maple I get the answer $(x+2)^3$ and I was just wondering what the steps were to get to this solution.
| \begin{align*}
x^3 - 1 &= (x - 1)(x^2 + x + 1)\\[6pt]
&\equiv (x + 2)(x^2 + 4x + 4) \pmod{3}\\[6pt]
&\equiv (x + 2)(x + 2)^2 \pmod{3}\\[6pt]
&\equiv (x + 2)^3 \pmod{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
maximum value of expression $6bc+6abc+2ab+3ac$ If $a,b,c>0$ and $a+2b+3c=15,$ then finding maximum value of $6bc+6abc+2ab+3ac$ is
with the help of AM - GM inequality
$4ab\leq (a+b)^2$ and $4bc \leq (b+c)^2$ and $\displaystyle 4ca \leq (c+a)^2$
and $27(abc)\leq (a+b+c)^3$
could some help me, thanks
| Let $a=5x$, $2b=5y$ and $3c=5z$.
Hence, $x+y+z=3$ and by AM-GM $bc+6abc+2ab+3ac=$
$$=25(xy+xz+yz)+ 125xyz\leq 25 \frac{(x+y+z)^2}{3} + 125 \left(\frac{x+y+z}{3}\right)^3=200.$$
The equality occurs for $x=y=z=1$, which says that the answer is $200$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.