Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Determine the limiting behaviour of $\lim_{x \to \infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$ Determine the limiting behaviour of $\lim_{x \to \infty}{\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$
Used L'Hopitals to get $\;\dfrac{(x^6+1)^{\frac{2}{3}}}{x^2 \sqrt{x^4+1}}$ but not sure what more i can do after that.
|
Dividing numerator and denominator by $x^2$ gives
$$\lim_{x \to +\infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}=\lim_{x\to+\infty} \frac{\sqrt{1+x^{-4}}}{\sqrt[3]{1+x^{-6}}}=\frac{\sqrt{1}}{\sqrt[3]{1}}=1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/367060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
}
|
Find expansion around $x_0=0$ into power series and find a radius of convergence My task is as in the topic, I've given function $$f(x)=\frac{1}{1+x+x^2+x^3}$$
My solution is following (when $|x|<1$):$$\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)+(x^2+1)}=\frac{1}{1-(-x)}\cdot\frac{1}{1-(-x^2)}=$$$$=\sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k$$ Now I try to calculate it the following way:
\begin{align}
& {}\qquad \sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k \\[8pt]
& =(-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+\cdots)\cdot(-x^2+x^4-x^6+x^8-x^{10}+\cdots) \\[8pt]
& =x^3-x^4+0 \cdot x^5+0 \cdot x^6 +x^7-x^8+0 \cdot x^9 +0 \cdot x^{10} +x^{11}+\cdots
\end{align}
And now I conclude that it is equal to $\sum_{k=0}^{\infty}(x^{3+4 \cdot k}-x^{4+4 \cdot k})$ ($|x|<1$)
Is it correct? Are there any faster ways to solve that types of tasks? Any hints will be appreciated, thanks in advance.
|
Let $x\ne 1$. By the usual formula for the sum of a finite geometric series, we have $1+x+x^2+x^3=\frac{1-x^4}{1-x}$. So your expression is equal to $\frac{1-x}{1-x^4}$.
Expand $\frac{1}{1-x^4}$ in a power series, multiply by $1-x$. Unless we are operating purely formaly, we will need to assume that $|x|\lt 1$.
For details, note that $\frac{1}{1-t}=1+t+t^2+t^3+\cdots$.
Substituting $x^4$ for $t$, we get that our expression is equal to
$$(1-x)(1+x^4+x^8+x^{12}+\cdots).$$
Now multiply through by $1-x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/369435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Show that $1$ and $2$ are zeros of the following polynomial
Show that $1$ and $2$ are zeros of the polynomial $P(x)=x^4-2x^3+5x^2-16x+12$ and
hence that $(x-1)(x-2)$ is a factor of $P(x)$
|
The is an application of the Factor Theorem.
If $\operatorname{f}(x)$ is a polynomial and $\operatorname{f}(p)=0$, then $x-p$ is a factor of $\operatorname{f}(x)$. In particular: if $\operatorname{f}(1)=0$ then $x-1$ is a factor of $\operatorname{f}(x)$ and if $\operatorname{f}(2)=0$ then $x-2$ is a factor of $\operatorname{f}(x)$. Hence to show that $\operatorname{f}(x)$ has $(x-1)(x-2)$ as a factor, we must show that both $\operatorname{f}(1)=0$ and $\operatorname{f}(2)=0$. Notice that:
\begin{array}{ccccc}
\operatorname{f}(1) &=& 1^4 - 2 \times 1^3 + 5 \times 1^2 - 16 \times 1 +12 &=& 1 - 2 + 5 - 16 + 12 &=& 0 \\
\operatorname{f}(2) &=& 2^4 - 2 \times 2^3 + 5 \times 2^2 - 16 \times 2 +12 &=& 16 - 16 + 20 - 32 + 12 &=& 0
\end{array}
It follows that both $x-1$ and $x-2$ divide $\operatorname{f}(x)$ and so their product $(x-1)(x-2)$ divides $\operatorname{f}(x)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/369562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Find real solutions of polynomial How with you find the solutions to the following:
$\sqrt{3x+10}-\sqrt{x+2}=2$
This is what I tried so far:
$(\sqrt{3x+10}-\sqrt{x+2})^2=2$
$(3x+10)+(x+2)-2\sqrt{(x+2)(3x+10)}=2$
$2x+5-\sqrt{3x^2+16x+20}=0$
No I do not know where to go from here...
|
Hint:
$2x+5-\sqrt{3x^2+16x+20}=0\iff2x+5=\sqrt{3x^2+16x+20} \Longrightarrow\\(2x+5)^2=3x^2+16x+20.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/373382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How show that $\dfrac{a^3}{3}\ge\int_{0}^{a}|F(x)-x|^2dx$ Let $F(x)$ be nonnegative and integrable on $[0,a]$ and such that
$$\left(\int_{0}^{t}F(x)dx\right)^2\ge\int_{0}^{t}F^3(x)dx$$
for every $t$ in $[0,a]$,prove or disprove the conjecture:
$$\dfrac{a^3}{3}\ge\int_{0}^{a}|F(x)-x|^2dx$$
This Problem from SIAM problem 78-18,I consider some time,but I failure it.
|
For simplicity let us write $F = F(x)$ and $G_t = \int_0^t F dx$.
Given $\displaystyle {G_t}^2 \ge \int_0^t F^3 dx$, we need to show $\displaystyle \dfrac{a^3}{3} \ge \int_0^a \mid F - x \mid^2 dx \tag{1}$.
Now, RHS $= \displaystyle \int_0^a \left( F - x \right)^2 dx = \int_0^a F^2 dx - \int_0^a 2x F dx + \frac{a^3}{3}$.
So we need to show $$\displaystyle \color{blue}{2\int_0^a x F dx} \ge \color{red}{\int_0^a F^2 dx} \tag{2}$$
By Hölder's inequality, we have: $$G_t = \int_0^t F dx \le \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot \left( \int_0^t 1 dx \right)^{\frac{2}{3}} = \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot t^{\frac{2}{3}} \le {G_t}^{\frac{2}{3}}\cdot t^{\frac{2}{3}} \quad \implies \sqrt{G_t} \le t $$
So we have: $\displaystyle \int_0^t x \cdot F dx \ge \int_0^t \sqrt{G_x}\cdot F dx = \frac{2}{3} {G_t}^{\frac{3}{2}}$ and $\color{blue}{\displaystyle 2\int_0^a x F dx \ge \frac{4}{3} {G_a}^{\frac{3}{2}}}$
Also by Hölder, we have:
$$\int_0^t F^2 dx = \int_0^t F^{\frac{3}{2}}\cdot F^{\frac{1}{2}} dx \le \left( \int_0^t F^3 dx \right)^{\frac{1}{2}} \cdot \left( \int_0^t F dx \right)^{\frac{1}{2}} \le G_t \sqrt{G_t} = {G_t}^{\frac{3}{2}} $$ and $\color{red}{\displaystyle \int_0^a F^2 dx \le {G_a}^{\frac{3}{2}}} $. Thus (2) holds, and hence (1).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/374172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
}
|
Finding the number of ordered pairs that satisfies an equation How many number of ordered pairs of positive integer $(a,d)$ satisfies:
$$\frac{1}{\frac{1}{a+2d}-\frac{1}{a+3d}}-\frac{1}{\frac{1}{a}-\frac{1}{a+d}}=2012$$
This question is too complicated. Like I tried to multiply by the common denominator, but it's not going anywhere since it's very complex. I tried to put it on wolframalpha and my ti89 they don't return me with anything.
Any clues?
Thanks!
|
The equation simplifies as
\begin{align}
&\frac{1}{\frac{1}{a+2d}-\frac{1}{a+3d}}-\frac{1}{\frac{1}{a}-\frac{1}{a+d}}=\frac{1}{\frac{d}{(a+2d)(a+3d)}}-\frac{1}{\frac{d}{a(a+d)}}\\ &=\frac{(a+2d)(a+3d)}{d}-\frac{a(a+d)}{d}\\&=\frac{a^2+5ad+6d^2-a^2-ad}{d}\\&=\frac{4ad+6d^2}{d}\\&=4a+6d\\&=2012
\end{align}
Thus, $2a+3d=1006$. Then, we can take this equation by modulo $2$, getting $d\equiv 0\ (mod \ 2)$, thus $d$ is even.
Now let $d=2x$, thus $x\in\mathbb{N}$. Taking the equation in modulo $3$, we get $2a\equiv 1\ (mod \ 3)$, thus $a\equiv 2\ (mod\ 3)$. Now, let $a=3y-1$, thus $y\in\mathbb{N}$.
By doing substitution, we get:
$$2(3y-1)+3(2x)=1006\implies 6y-2+6x=1006\implies x+y=168$$
Thus, there're $\boxed{167}$ ordered pairs where $a,d\in\mathbb{N}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/374240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Probability of Population A population consists of $50$% men and $50$% women of a population of $50$ people. A simple random sample (draws at random without replacement) of $4$ people is chosen. Find the chance that in the sample
i) the fourth person is a woman?
ii) The third person is a woman, given that the first person and fourth person are both men?
My attempt:
i) $P(4\text{th Women})=\frac{\dbinom{25}4}{\dbinom{50}4}=\;$?
ii) Conditionality $P(3\text{rd person Women}\mid\text{The }1\text{st and }4\text{th are both men})$?? Way Forward?
|
We can split it into, the probability that the first three were all women, $2$ women, $1$ woman, or no women. So
$$P(4\text{th Woman})=\frac{\binom{25}{3}}{\binom{50}{3}}\frac{22}{47}+\binom{3}{1}\frac{\binom{25}{2}\binom{25}{1}}{\binom{50}{2}\binom{48}{1}}\frac{23}{47}+\binom{3}{2}\frac{\binom{25}{1}\binom{25}{2}}{\binom{50}{1}\binom{49}{2}}\frac{24}{47}+\frac{\binom{25}{3}}{\binom{50}{3}}\frac{25}{47}$$
For the second part:
$$\begin{align*}
P(3\text{rd Woman} | \text{$1$st and $4$th Men})&=\frac{P(\text{$3$rd Woman and $1$st and $4$th Men})}{P(\text{$1$st and $4$th Men})}\\
&=\frac{25\cdot24\cdot25\cdot23+25\cdot25\cdot24\cdot24}{25\cdot24\cdot23\cdot22+\binom{2}{1}25\cdot24\cdot25\cdot23+25\cdot25\cdot24\cdot24}
\end{align*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/376990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Finding the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$
Find the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$.
Please brief about the concept behind this to solve such problems. Thanks.
|
2^100 + 3^100 + 4^100 + 5^100/7
= (2^3)^33 *2 + (3^3)^33 *3 + (4^3)^33 *3 + (5^3)^33 *3 / 7
= 2*8^33 + 3*27^33 + 4*64^33 + 5*125^33 / 7
= 2*1^33 + 3*-1^33 + 4*1^33 + 5*-1^33 / 7
= 2*1 + 3*-1 + 4*1 + 5*-1 / 7
= 2 - 3 + 4 -5 / 7
= -2/7
= 7-2
= 5 remainder
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/377378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
}
|
Calculus Reduction Formula
For any integer $k > 0$, show the reduction formula
$$\int^{2}_{-2} x^{2k} \sqrt{4-x^2} \, dx = C_k \int^{2}_{-2} x^{2k-2} \sqrt{4-x^2} \, dx$$
for some constant $C_{k}$.
(original image)
I thought this would be fairly straightforward but im a little confused. Do I start out by doing a trig substitution?
|
It took a little time to check through this, but here is an approach using the substitution I'd proposed in the comments. Applying $\ \sin\theta = \frac{x}{2}$ , we have
$$\int_{-2}^{2} x^{2k} \ \sqrt{4-x^2} \ dx \ \rightarrow \ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2^{2k} \cdot \sin^{2k}\theta \cdot (2 \cos\theta) \cdot (2 \cos\theta \ d\theta )$$
$$= \ 4^{k+1}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2k}\theta \ \cos^2\theta \ d\theta \ = \ 4^{k+1}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2k} \theta \ - \ \sin^{2k+2} \theta \ d\theta \ ,$$
having applied the Pythagorean Identity in this last stage. I will use the result (which I won't derive here)
$$ \ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2k}\theta \ d\theta \ = \ \frac{\sqrt{\pi} \cdot \Gamma(k+\frac{1}{2})}{\Gamma(k+1)} \ = \ \frac{\sqrt{\pi} \cdot (\frac{2k-1}{2} \cdot \ldots \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi})}{k!} $$
$$= \ \frac{ ([2k-1] \cdot \ldots \cdot 3 \cdot 1)}{2^k \cdot k!} \cdot \pi $$
$$\Rightarrow \ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2k} \theta \ - \ \sin^{2k+2} \theta \ d\theta \ = \ [ \frac{ ([2k-1] \cdot \ldots \cdot 3 \cdot 1)}{2^k \cdot k!} - \frac{ ([2k+1] \cdot \ldots \cdot 3 \cdot 1)}{2^{k+1} \cdot (k+1)!} ] \cdot \pi$$
$$= \ \frac{[ \ 2(k+1) \ - \ (2k+1) \ ] \cdot ([2k-1] \cdot \ldots \cdot 3 \cdot 1)}{2^{k+1} \cdot (k+1)!} \cdot \pi \ = \ \frac{ [2k-1] \cdot \ldots \cdot 3 \cdot 1}{2^{k+1} \cdot (k+1)!} \cdot \pi \ .$$
Our original integral is then
$$\int_{-2}^{2} x^{2k} \ \sqrt{4-x^2} \ dx \ = \ 4^{k+1} \cdot \frac{ [2k-1] \cdot \ldots \cdot 3 \cdot 1}{2^{k+1} \cdot \ (k+1)!} \cdot \pi $$
$$\Rightarrow \ C_k \ = \ \frac{4^{k+1} \cdot \frac{ [2k-1] \ \cdot \ [2k-3] \ \cdot \ \ldots \ \cdot \ 3 \ \cdot \ 1}{2^{k+1} \cdot \ (k+1)!} \cdot \pi}{4^k \cdot \frac{ [2k-3] \ \cdot \ \ldots \ \cdot \ 3 \ \cdot 1}{2^k \ \cdot \ k!} \cdot \pi} \ = \ \frac{4 \cdot (2k-1)}{2 \cdot (k+1)} \ = \ \frac{2 (2k-1)}{k+1} \ . $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/380025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Find the point of intersection of the line and surface I have an odd problem with no solution. I am completely lost on how to solve this.
Problem:
Find the coordinates of the point(s) of intersection of the line $x = 1+t$, $y = 2+3t$, $z = 1-t$ and the surface $z = x^2 +2y^2$
Attempt:
$(1) \ x = 1+t$
$(2) \ y = 2+3t$
$(3)\ z = 1-t$
Subbing in $(1), (2), (3)$ into the surface I have
$1-t = (1+t)^2 +2(2+3t)^2$
Solving I get:
$19t^2+27t+8=0$
At this point I need to solve for $t$
It appears to be a quadratic equation.
\begin{array}{*{20}c} {t = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}
\begin{array}{*{20}c} {t = \frac{{ - 27 \pm \sqrt {(27)^2 - 4(19)(8)} }}{{2(19)}}} \end{array}
\begin{array}{*{20}c} {t = \frac{{ - 27 \pm \sqrt {121} }}{{38}}} \end{array}
\begin{array}{*{20}c} \frac{{ - 27 \pm 11 }}{{38}} \end{array}
and
\begin{array}{*{20}c}\frac{{ - 27 + 11 }}{{38}} and \frac{{ - 27 - 11 }}{{38}} \end{array}
\begin{array}{*{20}c} t= \frac{{ 8 }}{{19}} \end{array} and
\begin{array}{*{20}c} t= -1 \end{array}
Then just plug these $t$ values into the parametric equations to get the points?
$(1) \ x = 1+t$ \begin{array}{*{20}c} x= 1 + \frac{{ 8 }}{{19}} \end{array}
$(2) \ y = 2+3t$ \begin{array}{*{20}c} x= 2 + 3 \frac{{ 8 }}{{19}} \end{array}
$(3) \ z = 1-t$ \begin{array}{*{20}c} x= 1 - \frac{{ 8 }}{{19}} \end{array}
Do the same for $t= -1$?
for the points $( 27/19, 26/19, 11/19)$ and $( 0, -1,2)$
Is this correct?
|
Your method is entirely correct, but as noted in the comment above, you may pick up unwanted solutions. Your point $(0, -1, 2)$ is on the surface, but this is not the case for $(27/19, 26/19, 11/19)$. (This point doesn't satisfy the equation of the surface. So if you double check your computations, and it results in the same coordinates, you can "throw it out," and keep $(0, -1, 2)$.
ADDED The root $t =\dfrac{8}{19}$ should be $t = \left(-\dfrac{8}{19}\right)$. Try computing the coordinates with the negative root $t = -\dfrac{8}{19}$, and see what this brings. Perhaps the point with the modified coordinates will satisfy the equation of the surface!
For $t = \left(-\dfrac{8}{19}\right)$, I get the point $\left(\dfrac{11}{19}, \dfrac{14}{19}, \dfrac{27}{19}\right)$. We need to simply check if it satisfies the equation of the surface. And so it does! (See another confirmation, below, thanks to @RecklessReckoner).
Therefore, we have two solutions!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/380576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
How to find the number of real roots of the given equation?
The number of real roots of the equation $$2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}$$ is
(A) $0$, (B) $1$, (C) $2$, (D) infinitely many.
Trial: $$\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-1} \left( \frac{2^x+2^{-x}}{2} \right)\end{align}$$ Then I can't proceed.
|
Notice that $2^0+2^0=2$ and that $2^x+2^{-x}> 2$ for $x\ne 0$. Then consider that $2\cos(0)=2$ and that $2\cos\left(\frac{x^2+x}{6}\right)\le2$ for all $x$. What can we conclude?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/380896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
}
|
Solve for $x$ a trigonometric equation I want to solve for $x$
$$ {{2}^{{{\sin }^{4}}x-{{\cos }^{2}}x}}-{{2}^{{{\cos }^{4}}x-{{\sin }^{2}}x}}=\cos 2x $$
but I don't know how to start. Replacing $\sin x$ or $\cos x$ by $y$ led me nowhere because of the right side.
One of the solutions I've found is $x=\pi/4$ but there could be more solutions though.
|
Putting $\cos^2x=a,$
$\sin^4x-\cos^2x=(1-a)^2-a=a^2-3a+1$ and $\cos^4x-\sin^2x=a^2-(1-a)=a^2+a-1$
So we get, $2^{a^2-3a+1}-2^{a^2+a-1}=2a-1$
or, $2^{a^2-3a+1}(1-2^{4a-2})=2a-1$
If $2a-1>0,$ $4a-2>0\implies 2^{4a-2}>2^0=1\implies $ the left hand side is $<0$
Similarly, if $2a-1<0$ the right hand side is $>0$
If $2a-1=0,$ both sides become $0$
$\implies 2\cos^2x-1=0\iff \cos2x=0\implies 2x=(2n+1)\frac{\pi}2$ where $n$ is any integer
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/381233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 2
}
|
Proving $\frac{x}{y} +\frac{y}{z} + \frac{z}{x} \ge 3$ for positive $x,y,z$
Suppose $x,y,z$ are real positive numbers, prove that:
$$\dfrac{x}{y} +\dfrac{y}{z} + \dfrac{z}{x} \ge 3$$
with equality when $x=y=z$.
Can someone help me find an easier solution?
I started with assume only two are equal, without loss of generality, $x=y$. Then we get
$$1 + \dfrac{x}{z} + \dfrac{z}{x}$$
$$1 + \dfrac{x^2 + z^2}{xz} = 1+ \dfrac{(x-z)^2 + 2xz}{xz} = 1 + 2 + \dfrac{(x-z)^2}{xz} \ge 3$$
To prove the general result I considered using a derivative with respect to $y$ to show that, starting with $x=y$ and increasing $y$ by a tiny amount, will show the equation's derivative as positive. I haven't gotten this to work yet. Any ideas?
|
Divide through by 3. From AM-GM, we have for any positive $a,b,c$
$$\frac{a+b+c}{3} \ge (abc)^{\frac{1}{3}}$$
With $a=\frac{x}{y}, b=...$, this becomes:
$$\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} \ge \left( \frac{x}{y} \frac{y}{z} \frac{z}{x} \right) ^{\frac{1}{3}}=1.$$
Now multiply by 3 again to get the answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/381673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Find all function $f:\mathbb{R}\mapsto\mathbb{R}$ such that $f(x^2+y^2)=f(x+y)f(x-y)$. Find all function $f:\mathbb{R}\mapsto\mathbb{R}$ such that $f(x^2+y^2)=f(x+y)f(x-y)$.
Some solutions I found are $f\equiv0,f\equiv1$, $f(x)=0$ if $x\neq0$ and $f(x)=1$ if $x=0$.
|
let $x=y=0 \to f(0)=f^2(0) \to f(0)=0$ or $ f(0)=1 $,
case I:$f(0)=0$
let $x-y=0 \to f(2x^2)=f(2x)*f(0)=0 \to f(x)=0$ when $x \ge 0 $,
let $x+y=u,x-y=v \to f(\dfrac{u^2+v^2}{2})=f(u)f(v), $if $v=c <0 ,f(c) \not=0$, $ f(u)=\dfrac{f(\dfrac{u^2+c^2}{2})}{f(c)}=f(-u) \to f(x)=0 $ when $x<0$, contradict. so $f(x)=0$ for all $x$.
EDIT: following is 3rd version and I think it is OK now.
case II: $f(0)=1 \to f(x^2)=(f(x))^2$ and $f(x)=f(\dfrac{x^2}{2})$ and $f(x)=f(-x)$
for any $x>\sqrt{2}$ we can have $ f(x)=(f(\sqrt{x}))^2=(f(x^{\frac{1}{2n}}))^{2n} $, when n is big enough, we have $x^{\frac{1}{2n}} < \sqrt{2}$
for any $x<\sqrt{2}$ we have $f(x)= f(\dfrac{x^2}{2})$, so $\dfrac{x^2}{2}<1$
for any $x<1 \to f(x)=f(\dfrac{x^2}{2})=f(\dfrac{x^{2n}}{2^{2n+1}})$, when n is big enough ,we have $f(x)=C$ when $x \to 0$
that gives for any $x>0, f(x)=C$, and $C=C^2 \to C=0 $ or $ C=1$
if $f(0)$ is not continue with $f(x)$ when $x=0$, we have $f(x)=0$ when $x\not=0$,or $f(x)=1$
so it proof that the OP's guess. QED
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/381724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
Showing that $\mathbb Q(\sqrt{17})$ has class number 1 Let $K=\mathbb Q(\sqrt{d})$ with $d=17$. The Minkowski-Bound is $\frac{1}{2}\sqrt{17}<\frac{1}{2}\frac{9}{2}=2.25<3$.
The ideal $(2)$ splits, since $d\equiv 1$ mod $8$. So we get $(2)=(2,\frac{1+\sqrt{d}}{2})(2,\frac{1-\sqrt{d}}{2})$ and $(2,\frac{1\pm\sqrt{d}}{2})$ are two ideals of norm $2$.
Now if we can show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal ideals, then we know that every ideal class contains a principal ideal, which shows that the class number is $1$.
But how can we show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal?
|
The quadratic integer $n=(1\pm\sqrt{17})/2$ has minimal polynomial $n^2-n-4$, so the ideals
$\left(2,\dfrac{1\pm\sqrt{17}}{4}\right)$
will be principal if natural integers $x,y$ exist such that
$|x^2-xy-4y^2|=2.$
As this condition holds when $x=2,y=1$ we are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/382188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
}
|
Polar equation representation of an exponential spiral involving $ \arctan(z) $ An exponential spiral is formulated as:
$$ \arctan(z) = \ln\left(\sqrt{1+z^2}\right)+\ln(x)+C $$ with: $$z=y/x$$
Represent the equation in polar coordinates and show $y' = \tan(\pi/4 + \theta)$.
Substituting in $ x = r\cos(\theta) $ and $ y = r\sin(\theta) $ I get:
$$ \arctan(z) = \ln\left(\sqrt{1+\frac{r^2\sin^2(\theta)}{r^2\cos^2(\theta)}}\,\right) + \ln(r\cos(\theta)) + C $$
$$ \arctan(z) = \ln\left(\sqrt{1+\tan^2(\theta)}\,\right) + \ln(r\cos(\theta)) + C $$
I'm not sure how to do this problem. I get stuck here.
|
If $x = r\cos\theta, y = r\sin\theta$, assuming we are in the first/fourth quadrant $\cos\theta >0$, otherwise your equation $\ln x$ would not make sense. Then the polar coordinates can be written as:
$$
r = \sqrt{x^2 + y^2}, \text{ and } \theta = \arctan\frac{y}{x}\tag{1}
$$
You have arrived at
$$ \arctan\frac{y}{x} = \ln\sqrt{1+\tan^2\theta}\, + \ln(r\cos\theta) + C \tag{2}$$
Plugging (1) into (2), exploiting the trigonometry identity $1+\tan^2\theta = \sec^2\theta$:
$$
\theta = \ln\sec\theta + \ln(r\cos\theta) + C
$$
which is
$$
\theta = \ln(\sec\theta\cdot r\cos\theta) +C = \ln r +C
$$
hence:
$$
r = c e^\theta \tag{3}
$$
and this is the equation for the exponential spiral. Plugging (3) back into the original polar coordinate parametrization:
$$
x = ce^\theta \cos\theta, \text{ and } y = c e^\theta\sin\theta
$$
For the $y'$:
$$
y' = \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\sin\theta + \cos\theta}{\cos\theta - \sin\theta}
$$
To get a tangent out, simply exploiting the formula:
$$\begin{aligned}
\cos(\theta +\frac{\pi}{4}) = \cos\theta \cos\frac{\pi}{4}- \sin\theta\sin\frac{\pi}{4}
\\
\sin(\theta +\frac{\pi}{4}) = \sin\theta \cos\frac{\pi}{4} + \cos\theta\sin\frac{\pi}{4}
\end{aligned}
$$
Therefore:
$$
y' = \frac{\sin(\theta +\frac{\pi}{4}) }{\cos(\theta +\frac{\pi}{4}) } = \tan(\theta +\frac{\pi}{4})
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/382401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that if $a, b, c$ are positive odd integers, then $b^2 - 4ac$ cannot be a perfect square. Prove that if $a, b, c$ are positive odd integers, then $b^2 - 4ac$ cannot be a perfect square.
What I have done:
This has to either be done with contradiction or contraposition, I was thinking contradiction more likely.
|
Suppose $b^2-4ac$ is a square. Since $a,b,c$ are odd, $b^2-4ac$ must be an odd square. Hence $x=\frac{-b+\sqrt{b^2-4ac}}{2}$ is an integer. Since $x^2+bx+ac=0$, therefore $x^2+bx+ac\equiv 0\,(mod\,2)$. By FLT $x^2\equiv x\,(mod\,2)$. Hence $(1+b)x+ac\equiv 0\,(mod\,2)$. Since $2|1+b$. Thus $0+ac\equiv 0\,(mod\,2)$ (contradiction)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/382564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
}
|
Solving a set of recurrence relation: Solving a set of recurrence relation:
$a_n=3b_{n-1} , b_n=a_{n-1}-2b_{n-1}$
In addition, It's known that: $a_1=2, b_1=1$.
So i started by isolating $b_n \to b_n=3b_{n-2}-2b_{n-1}$
So i get the current equation: $x^2+2x-3=0 \to(x+3)(x-1)=0$, Which means $b(n)=A_1+A_2(-3)^n$
So what now? I found an equation with 2 parameters but only 1 relating b: $b_1=1$.
Can i use $a_1=2$ aswell on the found equation?
|
Define generating functions $A(z) = \sum_{n \ge 0} a_{n + 1} z^n$ and similarly $B(z)$. Write your recurrences as:
$$
\begin{align*}
a_{n + 1} &= 3 b_n \\
b_{n + 1} &= a_n - 2 b_n
\end{align*}
$$
By properties of generating functions, your recurrences translate to:
$$
\begin{align*}
\frac{A(z) - a_1}{z} &= 3 B(z) \\
\frac{B(z) - b_1}{z} &= A(z) - 2 B(z)
\end{align*}
$$
Thus:
$$
\begin{align*}
A(z) &= \frac{2 + 7 z}{1 + 2 z - 3 z^2}
= \frac{9}{4} \frac{1}{1 - z} - \frac{1}{4} \frac{1}{1 + 3 z} \\
B(z) &= \frac{1 + 2 z}{1 + 2 z - 3 z^2}
= \frac{3}{4} \frac{1}{1 - z} + \frac{1}{4} \frac{1}{1 + 3 z}
\end{align*}
$$
Everything in sight is a geometric series:
$$
\begin{align*}
a_{n + 1} &= \frac{9}{4} - \frac{1}{4} (-3)^n \\
b_{b + 1} &= \frac{3}{4} + \frac{1}{4} (-3)^n
\end{align*}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/383192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$ Find all real numbers $x$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac76$$
I have tried to fiddle with it as follows:
$$2^{3x} \cdot 6 +3^{3x} \cdot 6=12^x \cdot 7+18^x \cdot 7$$
$$ 3 \cdot 2^{3x+1}+ 2 \cdot 3^{3x+1}=7 \cdot 6^x(2^x+3^x)$$
Dividing both sides by $6$ gives us
$$2^{3x}+3^{3x}=7 \cdot 6^{x-1}(2^x+3^x)$$
Is this helpful? If so, how should I proceed form here? If not any hints would be greatly appreciated.
|
$\Rightarrow $ $6(8^x+27^x)=7(12^x+18^x)$
Divide by$12^x$
$\Rightarrow $$ 6((\frac{2}{3}) ^x+(\frac{3}{2})^{2x}) =7(1+(\frac{3}{2})^{x})$
$\Rightarrow $ $(\frac{2}{3}) ^x+(\frac{3}{2})^{2x}-\frac{7}{6}-(\frac{7}{6})(\frac{3}{2}) ^x=0$
We put :$t=(\frac{3}{2}) ^x$
$\Rightarrow $ $\frac{1}{t} +t^2 - \frac{7}{6}t-\frac{7}{6}=0$
Multiply by $t$
$\Rightarrow $ $ 1+t^3 - \frac{7}{6}t^2-\frac{7}{6} t=0$
We can see $-1$ is a solution of the equation
So:after division by $t+1$ we see
$ t^2 - \frac{13}{6}x+1=0$
$\triangle =(\frac{13}{6})^2 - 4=(\frac{5}{6}) ^2 $
So :
$t_1=\frac{\frac{13}{6}+\frac{5}{6}}{2}=\frac{3}{2} =(\frac{3}{2})^{x_1} $
$\Rightarrow $ $x_1=1$
And
$t_2=\frac{\frac{13}{6}-\frac{5}{6}}{2}=\frac{2}{3}=(\frac{3}{2})^{x_2} $
$\Rightarrow $ $x_2=-1$
Finally
$x=-1$ or $ x= 1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/384090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 3,
"answer_id": 2
}
|
Laplace equation and integral $$ \int_0^{2\pi} \frac{1+3 \sin{\phi}}{a^2-2ar \cos(\theta - \phi) + r^2 } d\phi$$
Help me plz ... I have tried to solve this. but I still don't know.
|
This integral may be evaluated using residue theory. In this case, convert the integral over $\phi$ to a contour integral over the unit circle in the complex plane, and evaluate the residues of the poles inside the unit circle. Let $z=e^{i \phi}$, then $d\phi = dz/(i z)$, $\cos{\phi} = (z+z^{-1})/2$, $\sin{\phi} = (z-z^{-1})/(2 i)$. Then, after some algebra, the above integral becomes
$$\frac{i}{a r} e^{i \theta} \oint_{|z|=1} \frac{dz}{z} \frac{z - i \frac{3}{2} (z^2-1)}{z^2 - \frac{a^2+r^2}{a r} e^{i \theta} z + e^{i 2 \theta}}$$
There are three poles for this integrand: $z=0$, $z=(a/r) e^{i \theta}$, and $z=(r/a) e^{i \theta}$. We must now evaluate the residues at these poles and evaluate whether these poles lie inside or outside the unit circle. These residues are found by using the formula
$$\text{Res}_{z=z_k} \left [ \frac{f(z)}{g(z)} \right ] = \frac{f(z_k)}{g'(z_k)}$$
which we may use because the poles are simple. The residues are as follows:
$$\begin{array}\\z=0 & -\frac{3}{2 a r} e^{-i \theta} \\ z=(a/r) e^{i \theta} & \frac{i}{a r} \frac{1-i \frac{3}{2} \left ( \frac{a}{r} e^{i \theta} - \frac{r}{a} e^{-i \theta}\right )}{\frac{a}{r}-\frac{r}{a}}\\ z=(r/a) e^{i \theta} & \frac{i}{a r} \frac{1-i \frac{3}{2} \left ( \frac{r}{a} e^{i \theta} - \frac{a}{r} e^{-i \theta}\right )}{\frac{r}{a}-\frac{a}{r}} \end{array} $$
The value of the integral is $i 2 \pi$ times the sum of the residues of the poles inside the unit circle. This means that we must evaluate the cases where $a \lt r$ and $a \gt r$ separately, as one or other of the poles is outside the unit circle according to these conditions. That is, when $a \lt r$, the sum of the relevant residues is
$$-\frac{3}{2 a r} e^{-i \theta} + \frac{i}{a r} \frac{1-i \frac{3}{2} \left ( \frac{a}{r} e^{i \theta} - \frac{r}{a} e^{-i \theta}\right )}{\frac{a}{r}-\frac{r}{a}}$$
and when $a \gt r$, the sum of the relevant residues is
$$-\frac{3}{2 a r} e^{-i \theta} + \frac{i}{a r} \frac{1-i \frac{3}{2} \left ( \frac{r}{a} e^{i \theta} - \frac{a}{r} e^{-i \theta}\right )}{\frac{r}{a}-\frac{a}{r}}$$
At this point, I will simply state the result of simplifying the above expressions and multiplying by $i 2 \pi$:
$$\int_0^{2 \pi} d\phi \frac{1+3 \sin{\phi}}{a^2-2 a r \cos{(\theta-\phi)} + r^2} = \begin{cases} \\ 2 \pi \frac{1+3 (a/r) \sin{\theta}}{r^2-a^2} & a \lt r\\ 2 \pi \frac{1+3 (r/a) \sin{\theta}}{a^2-r^2} & a \gt r\end{cases}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/385337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Circular determinant problem I'm stuck in this question:
How calculate this determinant ?
$$\Delta=\left|\begin{array}{cccccc}
1&2&3&\cdots&\cdots&n\\
n&1&2&\cdots&\cdots& n-1\\
n-1&n&1&\cdots&\cdots&n-2\\
\vdots &\ddots & \ddots&\ddots&&\vdots\\
\vdots &\ddots & \ddots&\ddots&\ddots&\vdots\\
2&3&4&\cdots&\cdots&1
\end{array}\right|$$
Thanks a lot.
|
I'll do the case $n=4$ and leave the general case for you.
It is useful to recall how elementary row operations effect the value of determinant, see for example
ProofWiki.
We get
$$
\begin{vmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
4 & 1 & 2 & 3
\end{vmatrix}
\overset{(1)}=
\begin{vmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
10 & 10 & 10 & 10
\end{vmatrix}=
10\begin{vmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
1 & 1 & 1 & 1
\end{vmatrix}
\overset{(2)}=
10\begin{vmatrix}
0 & 1 & 2 & 3 \\
0 & 1 & 2 &-1 \\
0 & 1 &-2 &-1 \\
1 & 1 & 1 & 1
\end{vmatrix}
\overset{(3)}=
10\begin{vmatrix}
0 & 0 & 0 & 4 \\
0 & 0 & 4 & 0 \\
0 & 1 &-2 &-1 \\
1 & 1 & 1 & 1
\end{vmatrix}
=10\cdot 4^2
$$
(1): added first three rows to the last one
(2): subtracted a multiple of the 4th row from the other rows
(3): subtracted 2nd row from the 1st one, 3rd row from the 2nd one
There are several possibilities how to see that the determinant of the last matrix is $4^2$. For example, you can use Laplace expansion several times, until you get $2\times2$ matrix; you can use the expression of the determinant using permutations (the only permutation which will give a non-zero summand gives $a_{14}a_{23}a_{32}a_{41}$ for this matrix); or you can exchange rows in such way that you get upper triangular matrix.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/386526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Evaluate a sum with binomial coefficients: $\sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$ $$\text{Find} \ \ \sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$$
I expanded the binomial coefficients within the sum and got $$\binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2$$
What does this equal to? I think this can help me evaluate the original sum.
|
First, use $k\binom{n\vphantom{1}}{k}=n\binom{n-1}{k-1}=n\binom{n-1}{n-k}$
$$
\sum_{k=0}^n(-1)^kk\binom{n}{k}^2
=n\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{n-1}{n-k}\tag{1}
$$
Next compute a generating function. The sum we want is the coefficient of $x^n$
$$
\begin{align}
n\sum_{m,k}(-1)^k\binom{n}{k}\binom{n-1}{m-k}x^m
&=n\sum_{m,k}(-1)^k\binom{n}{k}\binom{n-1}{m-k}x^{m-k}x^k\\
&=n\sum_k(-1)^k\binom{n}{k}(1+x)^{n-1}x^k\\
&=n(1+x)^{n-1}(1-x)^n\\
&=n\left(1-x^2\right)^{n-1}(1-x)\tag{2}
\end{align}
$$
The sum we want is the coefficient of $x^n$ in $(2)$:
$$
\begin{align}
\sum_{k=0}^n(-1)^kk\binom{n}{k}^2
&=\left\{\begin{array}{}
n\binom{n-1}{n/2}(-1)^{n/2}&\quad\text{if $n$ is even}\\[6pt]
n\binom{n-1}{(n-1)/2}(-1)^{(n+1)/2}&\quad\text{if $n$ is odd}
\end{array}\right.\\[6pt]
&=n\binom{n-1}{\lfloor n/2\rfloor}(-1)^{\lceil n/2\rceil}\tag{3}
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/387171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
How to calculate ellipse sector area *from a focus* How do you calculate the area of a sector of an ellipse when the angle of the sector is drawn from one of the focii? In other words, how to find the area swept out by the true anomaly?
There are some answers on the internet for when the sector is drawn from the center of the ellipse, but not from the focii.
|
Following the approach I used for
Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse,
parametrize the ellipse as:
$x(t)=a \cos (t)$
$y(t)=b \sin (t)$
with $a>b$ (the case $a<b$ is symmetric). Note that $t$ is neither the
central angle nor the focal angle. The foci are then on the x-axis
with $x=\pm\sqrt{a^2-b^2}$ and the top right quadrant of the ellipse
looks like this:
where $d = a \cos (t)-\sqrt{a^2-b^2}$
This gives us:
$\tan (\theta ) = \frac{b \sin (t)}{a \cos (t)-\sqrt{a^2-b^2}}$
Solving for t, we have:
$t(\theta) = \cos ^{-1}\left(\frac{a \sqrt{(a-b) (a+b)} \tan ^2(\theta
)+b^2 \sec (\theta )}{a^2 \tan ^2(\theta )+b^2}\right)$
provided that $\theta<\pi/2$ (the case $\theta>=\pi/2$ is similar,
though not symmetric, and is left as an exercise for the reader)
We can now reparametrize the ellipse using the focal angle:
$x(\theta )=a \cos (t(\theta ))$
$y(\theta )=b \sin (t(\theta ))$
or:
$x(\theta) = \frac{a \left(a \sqrt{(a-b) (a+b)} \tan ^2(\theta )+b^2
\sec (\theta )\right)}{a^2 \tan ^2(\theta )+b^2}$
$y(\theta) = b \sqrt{1-\frac{\left(a \sqrt{(a-b) (a+b)} \tan ^2(\theta
)+b^2 \sec (\theta )\right)^2}{\left(a^2 \tan ^2(\theta
)+b^2\right)^2}}$
and compute the radius squared for a given $\theta$ (we could also
compute the radius itself, but we won't need it):
$r(\theta )^2=x(\theta )^2+y(\theta )^2$
Substituting and simplifying, this gives us:
$r(\theta)^2 = \frac{a^2 \tan ^2(\theta ) \left(\left(a^4-a^2
b^2+b^4\right) \tan ^2(\theta )+2 b^4\right)+b^4 (a-b) (a+b) \sec
^2(\theta )+2 a b^2 ((a-b) (a+b))^{3/2} \tan ^2(\theta ) \sec (\theta
)+b^6}{\left(a^2 \tan ^2(\theta )+b^2\right)^2}$
Now we simply integrate $r(\theta)^2/2$ over $d\theta$:
$A(\theta) = \int_0^{\theta } \frac{r(\phi)^2}{2} \, d\phi$
After integrating and simplifying, we have $\text{A}(\theta)=$
$\frac{1}{2} \left(\theta \left(a^2+b^2\right)+b \left(a \cot
^{-1}\left(\frac{b \csc (\theta )}{\sqrt{a^2-b^2}}\right)-\frac{2 b
\sin (\theta ) \left(\left(b^2-a^2\right) \cos (\theta )+a \sqrt{(a-b)
(a+b)}\right)+a \left(\left(b^2-a^2\right) \cos (2 \theta
)+a^2+b^2\right) \tan ^{-1}\left(\frac{a \tan (\theta
)}{b}\right)}{\left(b^2-a^2\right) \cos (2 \theta
)+a^2+b^2}\right)\right)$
As a reminder, this only works when $a>b$ (though the other case is
symmetric) and $\theta<\pi/2$ (the other case is similar, but not
symmetric).
Details on how I worked this out:
https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-ellipse-focus.m
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/388134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
}
|
Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.
Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.($a^2+b^2$, is same $b^2+a^2$), $n \in \mathbb{N}.$
I have proved the above question which appeared in one of the Math-Olympiad. And I do know the solution. Sharing the question only because the question has a cute solution.
|
*
*Let $n = 4x^4$, we have:
$$n(n+1) = (4x^4)^2 + (2x^2)^2 = (4x^4-2x^2)^2 + (4x^3)^2$$
*Let $n = (u^2 + v^2)^2$, we have
$$\begin{align}
n(n+1) = & ((u^2 + v^2)^2)^2 + (u^2+v^2)^2\\
= & (u^4 - 2uv - v^4)^2 + (2uv^3-v^2+2u^3v+u^2)^2\\
= & (u^4 + 2uv - v^4)^2 + (2uv^3+v^2+2u^3v-u^2)^2
\end{align}$$
*Let $n = (x+y)^2 + (2xy)^2$, we finally have an example that $n$ is not a square:
$$\begin{align}
n(n+1) = & (4x^2y^2+2xy+y^2-x^2)^2 + (4x^2y+y+x)^2\\
= & (4x^2y^2+2xy-y^2+x^2)^2 + (4y^2x+y+x)^2
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/388501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
}
|
Derivative of Trig Functions (Intuition Help?) Looking for some intuition help here.
I have the following exercise and these are the steps I take:
$$
y = \sin\left(\frac{1}{x}\right)
$$
$$
u=\frac{1}{x}
$$
$$
y = \sin u,\;\;\frac{dy}{du} = \cos u= \cos\left(\frac{1}{x}\right)
$$
$$
u=x^{-1};\;\frac{du}{dx} =-x^{-2}=-\frac{1}{x^2}
$$
$$
\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=cos\frac{1}{x}\times-\frac{1}{x^2}
$$
This is incorrect but intuitively I want to multiply it this way.
$$
cos\frac{1}{x}\times-\frac{1}{x^2}=cos-\frac{1}{x^3}
$$
But the correct answer is:
$$
-\frac{cos-\frac{1}{x}}{x^2}
$$
Help me absorb the why so I can intuitively solve problems like these.
|
First, you are taking short cuts when you are writing your equations that simply confuse things. Also, LaTeX has \sin and \cos. Start over.
$$
\begin{aligned}
\text{Let}\quad f(x) &= \sin\frac{1}{x}.\\
\text{Let}\quad u &= \frac{1}{x}.\\
f(x) &= \sin u\\
\frac{df}{dx} &= \frac{df}{du} \frac{du}{dx} \quad\text{by the Chain rule}\\
\frac{df}{du} &= \cos u\\
\frac{du}{dx} &= -\frac{1}{x^2} \\
\frac{df}{dx} &= \frac{df}{du} \frac{du}{dx} \\
&= \cos u \cdot [-\frac{1}{x^2}] \\
&= \cos \frac{1}{x} \cdot [-\frac{1}{x^2}]\\
&= -\frac{1}{x^2} \cos \frac{1}{x}.
\end{aligned}
$$
By the way, you are panicking and falling into the "Universal Distributive Law of Freshmen"; you think that all operations distribute over each other, and all operations commute. This is what has caused legions of students to replace $\sqrt{1+x^2}$ with $1 + x$. You wrote:
$$
\cos\frac1x \cdot \frac{1}{x^2} = \cos\frac{1}{x^3}.
$$
Even though $\cos x$ looks like a product, it isn't. Had you been using a programming language, you'd see that COS(X)*Y is not the same as COS(X*Y). Try any of these cases with actual numbers and a pocket calculator; you'll see what I mean.
When you panic like that, always take a deep breath and start over. And, never write a function next to its derivative and equate them.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/389037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
$x$ as the shortest alternating sum of $1 \ldots n$ If I have an positive integer $x \in \mathbb{N}$ and I have $Z = \sum_{i = 0}^{n}{i}$ such that $Z \geq x$ and $Z - x \equiv 0 \bmod 2$ and $n$ is the smallest such integer it is possible to create and alternating sum from $1 \ldots n$ such that it equals $x$.
For example:
Say $x = 11$, then $Z = 15$ and $n = 5$ and the sum is $1 - 2 + 3 + 4 + 5$.
I see why the condition is necessary I don't see why it is sufficient.
|
Let $S(n)=\{x:x=\pm1 \pm2 \cdots \pm n\}$. Let $T(n)=\{x:|x|\le \frac{n(n+1)}{2}\}\cap\{x:x\equiv \frac{n(n+1)}{2}\pmod{2}\}$.
Claim: $S(n)=T(n)$ for all $n\in \mathbb{N}$.
Proof: Induction on $n$. $n=1,2$: clear
$S(n+1)\subseteq T(n+1)$: consider all $+$ or all $-$
$T(n+1)\subseteq S(n+1)$: $$[0,\frac{(n+1)(n+2)}{2}]\subseteq[-\frac{n(n+1)}{2}+n,\frac{n(n+1)}{2}+n]$$ and $$[-\frac{(n+1)(n+2)}{2},0]\subseteq[-\frac{n(n+1)}{2}-n,\frac{n(n+1)}{2}-n]$$
Now, let $ x\in T(n+1)$ and $x\equiv \frac{(n+1)(n+2)}{2}\pmod{2}$. By the above pair of inclusions, either $x+n$ or $x-n$ is in $T(n-1)$. But also $\frac{(n+1)(n+2)}{2}- n =\frac{n^2+3n+2- 2n}{2}\equiv \frac{n^2+n+2}{2}\equiv \frac{n(n+1)}{2} \pmod{2}$. [and $n\equiv -n\pmod{2}$]. Hence we're done by the inductive hypothesis.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/389429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists? Here is my candidate for
the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists.
I would be interested in seeing others.
$***$ Added after some comments:
I prove here by very elementary means that the limit exists.
Calling the limit "$e$" names it.
$***$
It only needs Bernoulli's inequality (BI)
in the form
$(1+x)^n \ge 1+nx$
for $x > -1$ and
$n$ a positive integer,
with equality only if
$x = 0$ or $n = 1$.
This is easily proved by induction:
It is true for $n=1$,
and $(1+x)^{n+1}
= (1+x)(1+x)^n
\ge (1+x)(1+nx)
= 1+(n+1)x+nx^2
\ge 1+(n+1)x
$.
(If $-1 < x < 0$,
if $1+nx \ge 0$,
the above proof goes through, and if
$1+nx < 0$,
$1+mx < 0$ for all $m \ge n$
so certainly $(1+x)^m > 1+mx$.)
This proof originally appeared in
N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563
and uses the
arithmetic-geometric mean inequality (AGMI)
in the form
$\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i$
(all $v_i$ positive) with equality if and only if all the $v_i$ are equal.
Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$.
We will prove that $a_n$ is an increasing sequence
and $b_n$ is an decreasing sequence.
Since $a_n < b_n$, this implies,
for any positive integers $n$ and $m$ with $m < n$
that $a_m < a_n < b_n < b_m$.
For $a_n$, consider n values of $1+1/n$ and $1$ value of $1$.
Since their sum is $n+2$
and their product is $(1+1/n)^n$,
by the AGMI,
$((n+2)/(n+1))^{n+1} > (1+1/n)^n$,
or $(1+1/(n+1))^{n+1} > (1+1/n)^n$,
or $a_{n+1} > a_n$.
For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$.
Since their sum is $n$
and their product is $(1-1/n)^n$,
by the AGMI,
$(n/(n+1))^{n+1} > (1-1/n)^n$
or $(1+1/n)^{n+1} < (1+1/(n-1))^n$,
or $b_n > b_{n+1}$.
Since $b_n-a_n
= (1+1/n)^{n+1} - (1+1/n)^n
= (1+1/n)^n(1/n)
=a_n/n
$
and every $a_n$ is less than any $b_n$
and $b_3 = (1+1/3)^4 = 256/81 < 4$,
$b_n-a_n < 4/n$,
so $b_n$ and $a_n$ converge to a common limit.
These proofs do not seem to be really elementary,
since they use the AGMI.
However, they use a special form of the AGMI,
where all but one of the values are the same,
and this will now be shown to be implied by BI,
and thus be truly elementary.
Suppose we have $n-1$ values of $u$ and $1$ value of $v$ with $u$ and $v$ positive.
The AGMI for these values is
$(((n-1)u+v)/n)^n \ge u^{n-1}v$
with equality if and only if $u = v$.
We will now show that this is implied by BI:
$(((n-1)u+v)/n)^n \ge u^{n-1}v$
is the same as
$(u+(v-u)/n)^n \ge u^n(v/u)$.
Dividing by $u^n$,
this is equivalent to
$(1+(v/u-1)/n)^n \ge v/u$.
By BI,
since $(v/u-1)/n > -1/n > -1$,
$(1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u$
with equality only if $n=1$ or $v/u-1 = 0$.
Thus BI implies this version of the AGMI.
|
Could this be ?
$$\begin{align} \frac{d}{dx} \ln x &= \lim_{h \to 0}\; \frac{\ln(x + h) − \ln x}{h} \\
&= \lim_{h \to 0} \;\frac{\ln(1 + \frac{h}{x})}{h} \\
&= \lim_{h \to 0} \;\ln\left((1 + \frac{h}{x})^{\frac{1}{h}}\right) \\
\end{align}$$
let $h=\frac xc$
$$\begin{align} \frac{d}{dx} \ln x &= \lim_{c \to \infty} \;\ln\;\left((1 + \frac{1}{c})^{\frac{c}{x}}\right) \\
&= \frac{1}{x} \;\lim_{c \to \infty}\; \ln\;\left((1 + \frac{1}{c})^{c}\right) \\
&= \frac{1}{x} \;\ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right)
\end{align}$$
Since $\frac{d}{dx} \ln x = 1/x$
$$\begin{align} \frac{1}{x} &= \frac{1}{x} \;\ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right) \\
1 &= \ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right) \\
e &= \lim_{c \to \infty}\;\left(1 + \frac{1}{c}\right)^{c}
\end{align}$$
That's not very rigorous but well...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/389793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
}
|
GgT, (polynomial) division and finite fields... Exercise: Let $f,g \in \mathbb{Z}_2[x]$ be the polynomials $f = x^6 + x^5 + x^4 + 1$ and $g = x^5 + x^4 + x^3 + 1$. Has the diophantic equation $f u + g u = x^4 + 1$ solutions $u,v \in \mathbb{Z}[x]$? If so, determine them all.
My Solution: First I observed that in $\mathbb{Z}_2$ the relations
\begin{align}
-x & = x \\
x + x & = 0 \\
x^n = x
\end{align}
hold. From this the polynomials $f,g$ reduce to $f = g = x + 1$, so they are the same and so $ggT(f,g) = x + 1$. Now $x^4 + 1 = x + 1$, so $u = 0, v = 1$ is a special solution. To get all solutions, I add solutions of the homogenous part $fu + gv = 0$, i.e. $fu = gv$ with my relations. But because $f = g$, this forces $u = v$, so I could choose for $u$ and $v$ some arbitrary polynomial. So all solutions $(u,v)$ are of the form
\begin{align}
u & = p \\
v & = 1 + p
\end{align}
for some polynomial $p \in \mathbb Z_2$.
Reference Solution: First it has to be checked if $x^4 + 1$ is a multiple of
the ggT of $f$ and $g$, then the equation has a solution (criterium for lineare diophantic equations). The euclidean algorithms yields the ggT. It is $f = (x+1)g + x^4 + x^3 + x^2 + x = (x+1)g + r_1$, $g = xr_1 + x^2 + 1$ and $r_1 = (x^2 + x)(x^2 + 1)$. So $ggT(f,g) = x^2 + 1$, and because of $(x^2+1)^2 = x^4 + 1$ the diophantic equation has a solution.
The general solution is given by the sum of a special solution and the general solution of the homogenous part $fu + gv = 0$. The special solution could be obtained by the euclidean algorithm (Lemma of Bezout). With $1 = -1$ we got
$$
x^2 + 1 = g + xr_1 = g + x(f + (x+1)g)) = xf + (x^2 + x + 1)g.
$$
One special solution for the representation of the ggT is therefore $(x, x^2 + x + 1)$, one special solution of the original equation $(x(x^2+1), (x^2+x+1)(x^2+1)) = (x^3 + x, x^4 + x^3 + x + 1)$.
For the homogenous part $fu = gv$ we divide by the ggT. It is $f = (x^4+1)(x^2+1)$ and $g = (x^3+x^2+1)(x^2+1)$, a solution $(u,v)$ thus has the property $x^2+x^2+1 | u$ and $x^4+1 | v$. Therefore $u = p \cdot (x^3 + x^2 + 1)$. Substitution yields $v = p\cdot (x^4 + 1)$ and furthermore every $u = p\cdot (x^3 + x^2 + 1)$ and $v = p\cdot (x^4 + 1)$ are solutions. Therefore $\{ (p\cdot (x^3+x^2+1), p\cdot (x^4+1)) : p \in \mathbb Z_2[x]\}$ is the solution set of the homogenous part. The solution set is therefore
$$
\{ (x^3 + x + p\cdot (x^3 + x^2 + 1), x^4 + x^3 + x + 1 + p\cdot (x^4+1)) : p \in \mathbb Z_2[x] \}.
$$
Now I understand the reference solution, it is the standard way of solving the problem. But my solution seems much simpler to mean, so I wonder why they didn't consider it, or maybe my solution is just wrong? But then why should it be wrong?
|
You are taking polynomials, so it is incorrect to say that $x^n = x$ in $\Bbb{Z}_{2}[x]$. Actually you have that
$$
a_{0} + a_{1} x + \dots + a_{n} x^{n}
=
b_{0} + b_{1} x + \dots + b_{n} x^{n}
$$
for two elements of $\Bbb{Z}_{2}[x]$
if and only if $a_i = b_i$ for all $i$.
Most likely you are mixing up polynomials and polynomial functions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/390390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$ I'm trying to solve this recurrence relation:
$$
a_n = \begin{cases}
0 & \mbox{for } n = 0 \\
5 & \mbox{for } n = 1 \\
6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
\end{cases}
$$
I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - 6x + 5x^2} + \frac{x^3}{(1-x)(1-6x+5x^2)} =
\frac{31x - 24x^2}{(x-1)(x-5)} + \frac{x^3}{(1-x)(x-1)(x-5)}
$$
(I'm not sure if that's right)
and its partial fractions decomposition looks like:
$$
A = \left(\frac{-7}{4} \cdot \frac{1}{x-1} - \frac{445}{4} \cdot \frac{1}{x-5}\right) +
\left( \frac{39}{16} \cdot \frac{1}{x-5} + \frac{3}{4} \cdot \frac{1}{(x-1)^2} - \frac{375}{16} \cdot \frac{1}{x-5} \right)
$$
(again - I'm not sure if it's ok)
I'm stuck here... From solutions I know that I should get:
$$
a_n = \frac{-21}{16} - \frac{1}{4}n + \frac{21}{16}5^n
$$
but I have no idea how it's solved... I hope somebody can help me (I spend more than 3h trying to solve this myself...)
|
$$a_0 =0,a_1=5,a_n=6a_{n-1} - 5a_{n-2} + 1, n > 1$$
$$f(x)=\sum_{n=0}^{\infty}a_nx^n=5x+\sum_{n=2}^{\infty}a_nx^n=$$
$$=5x+\sum_{n=2}^{\infty}(6a_{n-1} - 5a_{n-2} + 1)x^n=$$
$$=5x+6\sum_{n=2}^{\infty}a_{n-1}x^n-5\sum_{n=2}^{\infty}a_{n-2}x^n+\sum_{n=2}^{\infty}x^n=$$
$$=5x+6x\sum_{n=2}^{\infty}a_{n-1}x^{n-1}-5x^2\sum_{n=2}^{\infty}a_{n-2}x^{n-2}+\frac{x^2}{1-x}=$$
$$=5x+6xf(x)-5x^2f(x)+\frac{x^2}{1-x}$$ from above follow that g.f. is
$$f(x)=\frac{5x+\frac{x^2}{1-x}}{5x^2-6x+1}=\frac{5x-4x^2}{(1-x)(5x^2-5x-(x-1))}=$$
$$=\frac{x(5-4x)}{(1-x)(5x(x-1)-(x-1))}=\frac{x(5-4x)}{(1-x)(x-1)(5x-1)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/390644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Integrating $\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^n \, d\theta$, for $n=2$ and $n=3$ How do you integrate the following functions:
$$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^2 \, d\theta$$ and $$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^3 \, d\theta
$$
respectively?
Note: Initially, I tried integrating the function without the power and obtained the result below.
$$
\int \frac{\cos\theta}{1+\sin^2\theta} \, d\theta = \arctan(\sin\theta)+ C
$$
However, from here it is difficult to proceed. Integration by substitution doesn't seem to work.
How should I go on from here? Any pointers would be greatly appreciated.
|
Multiplying both numerator and denominator by $\sec^2 \theta$ yields
$$
\begin{aligned}
I_2=&\int\left(\frac{\cos \theta}{1+\sin ^{2} \theta}\right)^{2} d \theta\\=& \int \frac{\sec ^{2} \theta}{\left(\sec ^{2} \theta+\tan ^{2} \theta\right)^{2}} d \theta \\
=& \int \frac{d t}{\left(1+2 t^{2}\right)^{2}}, \text { where } t=\tan \theta
\end{aligned}
$$
$$
\begin{aligned}
I_{2} & =\int \frac{d t}{\left(1+2 t^{2}\right)^{2}}\\&=-\frac{1}{4}\int\frac{1}{t} d\left(\frac{1}{1+2 t^{2}}\right)\\
&=-\frac{1}{4 t\left(1+2 t^{2}\right)}-\frac{1}{4}\left(\frac{1}{t^{2}\left(1+2 t^{2}\right)} d t\right.\\
&=-\frac{1}{4 t\left(1+2 t^{2}\right)}-\frac{1}{4} \int\left(\frac{1}{t^{2}}-\frac{2}{1+2 t^{2}}\right) d t \\
&=-\frac{1}{4 t\left(1+2 t^{2}\right)}+\frac{1}{4 t}+\frac{1}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} t)+C \\
&=\frac{t}{2\left(2 t^{2}+1\right)}+\frac{1}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} t)+C \\
&=\frac{\tan \theta}{2\left(2 \tan ^{2} \theta+1\right)}+\frac{1}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} \tan \theta)+C
\end{aligned}
$$
Now let’s go further to $I_3$ by the substitution $\tan u=\sin \theta$.
$$\begin{aligned} \int \frac{\cos ^{3} \theta}{\left(1+\sin ^{2} \theta\right)^{3}} d \theta &=\int \frac{1-\tan ^{2} u}{\left(1+\tan ^{2} \theta\right)^{3}} \cdot \sec ^{2} u d u \\ &=\int \frac{1-\tan ^{2} u}{\sec ^{4} u} d u \\ &=\int\left(\cos ^{4} u-\sin ^{2} u \cos ^{2} u\right) d u \\ &=\int \cos ^{2} u\left(\cos ^{2} u-\sin ^{2} u\right) d u \\ &=\int \frac{1+\cos 2 u}{2} \cos 2 u d u \\ &=\frac{\sin 2 u}{4}+\frac{1}{2} \int \frac{1+\cos 4 u}{2} d u \\ &=\frac{\sin u \cos u}{2}+\frac{1}{4} u+\frac{1}{16} \sin 4 u+C \\ &=\frac{\sin u \cos u}{2}+\frac{u}{4}+\frac{\sin u \cos u \cos 2 u}{4}+C \\ &=\frac{u}{4}+\frac{\sin u \cos u}{4}(2+\cos 2 u)+C \\ & =\frac{1}{4} \tan ^{-1}(\sin \theta)+\frac{\sin \theta}{4\left(1+\sin ^{2} \theta\right)}\left(2+\frac{1-\sin ^{2} \theta}{1+\sin ^{2} \theta}\right)+C\\& =\frac{1}{4} \tan ^{-1}(\sin \theta)+\frac{\sin \theta\left(\sin ^{2} \theta+3\right)}{(\cos 2 \theta-3)^{2}}+C\end{aligned}$$
:|D Wish you enjoy my solution!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/391338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
Remainders problem What will be the reminder if $23^{23}+ 15^{23}$ is divided by $19$?
Someone did this way:
$15/19 = -4$ remainder and $23/19 = 4$ remainder So $(-4^{23}) + (4^{23}) =0$ but i didn't understand it
|
We have $23\equiv 4\pmod{19}$ and $15\equiv -4\pmod{19}$. Then for any odd positive integer $n$, we have $23^n\equiv 4^n \pmod{19}$ and $15^n\equiv (-4)^n \equiv (-1)^n4^n\equiv -(4^n)\pmod{19}$.
Add. We get $4^n -4^n\equiv 0 \pmod{19}$.
Added: We don't really need to go to $4$ and $-4$. For note that $23\equiv -15 \pmod{19}$. So if $n$ is any odd integer, then $23^n +15^n\equiv (-15)^n +15^n\equiv -15^n+15^n\equiv 0\pmod{19}$.
And we do not really need congruence notation. For recall that if $n$ is odd, then $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\cdots +b^{n-1}$. Let $a=23$ and $b=15$. We get that $23+15$ divides $23^{23}+15^{23}$. This yields the desired result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/391753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
What is the general equation of a cubic polynomial? I had this question:
"Find the cubic equation whose roots are the the squares of that of $x^3 + 2x + 1 = 0$"
and I kind of solved it. In that my answer was $x^3 - 4x^2 + 4x + 1$, but it was actually $x^3 + 4x^2 + 4x - 1 = 0$.
I took the general equation of a cubic equation, which was: $x^3 +bx^2/a + cx/a + d/a$.
Through simultaneous equations, I found what $b/a, c/a, d/a$ should equate to for my unknown cubic polynomial. Am I supposed to make $b/a, c/a, d/a$ all positive, then substitute it into the general formula?
Any help would be greatly appreciated, thanks.
|
Let $a,b,c$ be roots of $x^3 + 2x + 1 = 0$
Using Vieta's Formulas, $a+b+c=0,ab+bc+ca=2,abc=-1$
So, we need to find the cubic equation whose roots are $a^2,b^2,c^2$
i.e., $$(y-a^2)(y-b^2)(y-c^2)=0$$
$$\iff y^3-(a^2+b^2+c^2)y^2+(a^2b^2+b^2c^2+c^2a^2)y+a^2b^2c^2=0$$
Now, $a^2b^2c^2=(abc)^2=(-1)^2=1$
$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=0-2(2)=-4$
$a^2b^2+b^2c^2+c^2a^2=(ab)^2+(bc)^2+(ca)^2$
$=(ab+bc+ca)^2-2(ab\cdot bc+bc\cdot ca+ca\cdot ab)$
$=(ab+bc+ca)^2-2(a+b+c)abc=2^2-2\cdot0\cdot(-1)=4$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/393312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ if $a+b+c=0$ I'm stuck at this algebra problem, it seems to me that's what's provided doesn't even at all.
Provided: $$a+b+c=0$$
Find the value of: $$\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$
Like I'm not sure where to start, and the provided clue doesn't even make sense. There's no way I can think of to factor this big polynomial into a form like $a\times p+b\times q+c\times r=s$ where $p,q,r,s\in\mathbb{Z}$.
Thanks!
|
Note that:
$$\frac{a^2}{2a^2+bc}=\frac{a^2}{a^2+a(-b-c)+bc}=\frac{a^2}{(a-b)(a-c)}$$
This applies in the same way for:
$$\frac{b^2}{2b^2+ac}=\frac{b^2}{(b-c)(b-a)}\ \text{and}\ \frac{c^2}{2c^2+ab}=\frac{c^2}{(c-a)(c-b)}$$Therefore, the original equation is equal to:
\begin{align}
&\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-c)(b-a)}+\frac{c^2}{(c-a)(c-b)}
\\
\\
&=\frac{-a^2(b-c)-b^2(c-a)-c^2(a-b)}{(a-b)(b-c)(c-a)}
\\
\\
&=\frac{-a^2b-b^2c-c^2a+a^2c+b^2a+c^2b}{(a-b)(b-c)(c-a)}
\\
\\
&=\frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}
\\
\\
&=\boxed{1}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/393786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
How to show this inequality?
Show that $$-2 \le \cos \theta(\sin \theta+\sqrt{\sin^2 \theta +3})\le2$$
Trial: I know that $-\dfrac 1 2 \le \cos \theta\cdot\sin \theta \le \dfrac 1 2$ and $\sqrt 3\le\sqrt{\sin^2 \theta +3}\le2$. The problem looks simple to me but I am stuck to solve this. Please help. Thanks in advance.
|
Essentially, we want the extremum of
$$f(a) = a\sqrt{1-a^2} + a\sqrt{4-a^2}$$
We have
$$f'(a) = \sqrt{1-a^2} + \sqrt{4-a^2} - \dfrac{a^2}{\sqrt{1-a^2}} - \dfrac{a^2}{\sqrt{4-a^2}} = \dfrac{1-2a^2}{\sqrt{1-a^2}} + \dfrac{4-2a^2}{\sqrt{4-a^2}}$$
Let $a^2 = x$, we then want to solve for
\begin{align}\dfrac{1-2x}{\sqrt{1-x}} + \dfrac{4-2x}{\sqrt{4-x}} = 0 \implies (1-2x)^2(4-x) = 4(2-x)^2(1-x)\\
(-4x^3+20x^2-17x+4)-(-4x^3+20x^2-32x+16) = 0 \implies15x-12=0 \implies x = \dfrac45
\end{align}
Hence, $a^*= \pm \dfrac2{\sqrt5}$, which gives us $f(a^*) = \pm 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/394618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
}
|
quadratic equation precalculus from Stewart, Precalculus, 5th, p56, Q. 79
Find all real solutions of the equation
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{x^2-4}$$
my solution
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{(x+2)(x-2)}$$
$$(x+2)(x+5)=5(x-2)+28$$
$$x^2+2x-8=0$$
$$\dfrac{-2\pm\sqrt{4+32}}{2}$$
$$\dfrac{-2\pm6}{2}$$
$$x=-4\text{ or }2$$
official answer at the back of the book has only one real solution of $-4$
where did I go wrong?
|
You multiplied both sides by $(x-2)(x+2)$. If this is zero, you may introduce extra solutions, hence you need to check your final answer to see if you have any extraneous solutions. In this case, you do! For $x=2$, two of the three fractions in the original equation are undefined.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/395691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$? The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$
|
The square of this is $2x+2\sqrt{x^2-y^2}$. One could find this.
Regarding $\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$, the modulus of this number $35+18\sqrt{-3}$ is 2197, which is $13^3$. This means that we are looking for a number whose cube is $35+18\sqrt{-3}$. An example of such a number is $3.5 + 0.5\sqrt{-3}$, which we cube to get the expression of $35+18\sqrt{-3}$.
The conjucate gives $3.5-0.5\sqrt{-3}$ gives a cube of $35-18\sqrt{-3}$.
The process here is to multiply the two numbers together, to establish the modulus. Here, the product gives $2197$, a known cube. In general, it needs to be a cube to fall on an integer real. One could also divide the two numbers, in the form $\sqrt[3]{35 + 18i\sqrt{3}} / \sqrt[3]{35 - 18i\sqrt{3}}$, to get a value on the unit circle.
However, it is easier to trace around a circle of radius $\sqrt{13}$, by looking for solutions to $x^2 + 3y^2 = 52$ for solutions of the type $(x+y\sqrt{-3})/2$. One finds numbers like $7,3$, and then $5,3$ and $2,4$, all of which are likely suspects.
It's a similar method that i have employed in more complex integer systems.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/396915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 5
}
|
Fisher information of a Binomial distribution The Fisher information is defined as $\mathbb{E}\Bigg( \frac{d \log f(p,x)}{dp} \Bigg)^2$, where $f(p,x)={{n}\choose{x}} p^x (1-p)^{n-x}$ for a Binomial distribution. The derivative of the log-likelihood function is $L'(p,x) = \frac{x}{p} - \frac{n-x}{1-p}$. Now, to get the Fisher infomation we need to square it and take the expectation.
First, we know, that $\mathbb{E}X^2$ for $X \sim Bin(n,p)$ is $ n^2p^2 +np(1-p)$. Let's first focus on on the content of the paratheses.
$$
\begin{align}
\Bigg( \frac{x}{p} - \frac{n-x}{1-p} \Bigg)^2&=\frac{x^2-2nxp+n^2p^2}{p^2(1-p)^2}
\end{align}
$$
No mistake so far (I hope!).
\begin{align}
\mathbb{E}\Bigg( \frac{x}{p} - \frac{n-x}{1-p} \Bigg)^2 &= \sum_{x=0}^n \Bigg( \frac{x}{p} - \frac{n-x}{1-p} \Bigg)^2 {{n}\choose{x}} p^x (1-p)^{n-x} \\
&=\sum_{x=0}^n \Bigg( \frac{x^2-2nxp+n^2p^2}{p^2(1-p)^2} \Bigg) {{n}\choose{x}} p^x (1-p)^{n-x} \\
&= \frac{n^2p^2+np(1-p)-2n^2p^2+n^2p^2}{p^2(1-p)^2}\\
&=\frac{n}{p(1-p)}
\end{align}
The result should be $\frac{1}{p(1-p)} $ but I've been staring at this for a few hours incapable of getting a different answer. Please let me know whether I'm making any arithmetic mistakes.
|
Fisher information: $I_n(p) = nI(p)$, and $I(p)=-\mathbb{E_p}\Bigg( \frac{\partial^2 \log f(p,x)}{\partial p^2} \Bigg)$, where $f(p,x)={{1}\choose{x}} p^x (1-p)^{1-x}$ for a Binomial distribution. We start with $n=1$ as single trial to calculate $I(p)$, then get $I_n(p)$.
$\log f(p,x) = x \log p + (1-x) \log p$
$\frac {\partial \log f(p,X)}{\partial p} = \frac {X}{p} - \frac {1- X}{1 - p}$
$\frac {\partial^2 \log f(p,X)}{\partial p^2} = -\frac {X}{p^2} - \frac {1- X}{(1 - p)^2}$
$I(P) = -\mathbb{E_p}\Bigg( \frac{\partial^2 \log f(p,x)}{\partial p^2} \Bigg) = -\mathbb{E_p}\Bigg(-\frac {X}{p^2} - \frac {1- X}{(1 - p)^2}\Bigg) = \frac {p}{p^2} + \frac {1-p}{(1-p)^2} = \frac {1}{p} + \frac {1}{(1-p)} = \frac {1}{p(1-p)} $
As a result, $I_n(p) = n I(p) = \frac {n}{p(1-p)} $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/396982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 1
}
|
Getting rid of a floor function in the next expression:$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $, It is known x is odd. I was wondering if there's a way in which you can get rid of a floor function in the next expression:$$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $$ It is known x is odd.
|
Since $x$ is odd, $x=2n+1$ for some integer $n$, and therefore $(x-2)^2=(2n-1)^2=4n^2-4n+1$. Thus,
$$\left\lfloor\frac{(x-2)^2}4\right\rfloor=\left\lfloor n^2-n+\frac14\right\rfloor=n^2-n\;.$$
And $n=\dfrac{x-1}2$, so the expression reduces to
$$\left(\frac{x-1}2\right)^2-\frac{x-1}2=\frac{x^2-4x+3}4\;.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/397435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
For x < 5 what is the greatest value of x It can't be $5$. And it can't be $4.\overline{9}$ because that equals $5$. It looks like there is no solution... but surely there must be?
|
If $x<5$ then $2x<x+5$ so $x<\frac{x+5}{2}$. Similarly, $x<5$ means $x+5<5+5=10$ or $\frac{x+5}{2}<5$. So if $x<5$ we have $x<\frac{x+5}{2}<5$, and therefore there is a larger number, $\frac{x+5}{2}$ less than $5$.
Basically, the average of two different numbers must be strictly between those two numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/397566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 2
}
|
Value of series, Partialsum? given is the following series
$$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$$
And I need to find its value.
How can I start finding it?
Thanks for all
does the Telescop-Summing work here as well?:
$\sum_{n=1}^\infty \frac{1}{4n^2-1} $
now:
$\frac{1}{4n^2-1} = \frac{1}{2} * \frac{(2n+1)-(2n-1)}{(2n+1)(2n-1)} = \frac{1}{2} * ( \frac{1}{2n-1} - \frac{1}{2n+1})$
Now I have to "add the sum":
$\sum_{n=1}^\infty \frac{1}{4n^2-1} = \frac{1}{2}* [ \sum_{n=1}^\infty \frac{1}{2n-1} -
\sum_{n=1}^\infty \frac{1}{2n+1}] = \frac{1}{2} - \frac{1}{4n+2} $ And than for $n \to \infty$ it is $\frac{1}{2}$ ??
|
HINT:
$$\frac{2n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}=\frac1{n^2}-\frac1{(n+1)^2}$$
Can you recognize Telescoping Sum?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/398075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Show $60 \mid (a^4+59)$ if $\gcd(a,30)=1$
If $\gcd(a,30)=1$ then $60 \mid (a^4+59)$.
If $\gcd(a,30)=1$ then we would be trying to show $a^4\equiv 1 \mod{60}$ or $(a^2+1)(a+1)(a-1)\equiv 0 \mod{60}$. We know $a$ must be odd and so $(a+1)$ and $(a-1)$ are even so we at least have a factor of $4$ in $a^4-1$. Was thinking I could maybe try to show that there is also a factor of $3$ and $5$ necessarily giving that $a^4-1\equiv0 \mod{60}$.
Other things I was thinking was that as $Ord_n(a) \mid \phi(n)=\phi(60)=16$ that we just need to show that $Ord_n(a) \in \{1,2,4 \}$.
Any hints? I have the exam soon =/
|
We essentially want to show that $60 \vert (a^4-1)$. Since $\gcd(a,30) = 1$, we have
$$a \equiv \pm1, \pm 7, \pm 11, \pm 13, \pm 17, \pm 19, \pm 23, \pm 29 \pmod{60}$$
Hence,
$$a^2 \equiv \begin{cases}1 \pmod{60} & \text{if }a \equiv \pm1,\pm11,\pm19,\pm29\\ -11 \pmod{60} & \text{if }a \equiv \pm7,\pm13,\pm17,\pm23\end{cases}$$
which inturn gives us
$$a^4 \equiv 1 \pmod{60} \,\,\,\, \forall a \text{ such that }\gcd(a,30) = 1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/398656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Show that for all $a,b,c>0$, $\frac 1 {\sqrt[3]{(a+b)(b+c)(c+a)}}\geq\frac 3 {2(a+b+c)}$.
Show that for all $a,b,c>0$, $\displaystyle\frac 1 {\sqrt[3]{(a+b)(b+c)(c+a)}}\geq\frac 3 {2(a+b+c)}$.
I tried to cube the both sides, and expand it, but that'll be too troublesome, is there simpler ways? Thasnk you.
|
HINT:
Using AM-GM inequality,
$$\frac{a+b+b+c+c+a}3\ge \sqrt[3]{(a+b)(b+c)(c+a)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/400036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Sum : $\sum \sin \left( \frac{(2\lfloor \sqrt{kn} \rfloor +1)\pi}{2n} \right)$. Calculate : $$ \sum_{k=1}^{n-1} \sin \left( \frac{(2\lfloor \sqrt{kn} \rfloor +1)\pi}{2n} \right).$$
|
Lemma Summation by Pasts (1)
$$\sum_{k=a}^b f_k\Delta g_k=f_kg_k\Bigg|_{k=a}^{b+1}-\sum_{k=a}^b g_{k+1}\Delta f_k=f_{b+1}g_{b+1}-f_ag_a-\sum_{k=a}^b g_{k+1}\Delta f_k$$
$\displaystyle \text{with difference operator }\Delta :\qquad \Delta f_k=f_{k+1}-f_k$
My solution
$\text{Get }j=\left\lfloor \sqrt{kn}\right\rfloor\Rightarrow j\le \sqrt{kn}<j+1\Rightarrow \dfrac{j^2}{n}\le k<\dfrac{(j+1)^2}{n}$
Therefore if $\left\lfloor\dfrac{j^2}{n}\right\rfloor+1\le k\le \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor$ then $\sqrt{kn}=j$
Note:
$j=0\Rightarrow \left\lfloor\dfrac{j^2}{n}\right\rfloor+1=1$
$j=n-2\Rightarrow \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor=n-2$
$j=n-1\Rightarrow \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor=n>n-1$
So that sum became
\begin{align*}S&=\sum_{k=1}^{n-1}\sin\left(\dfrac{\left(2\lfloor\sqrt{kn}\rfloor+1\right)\pi}{2n}\right)\\ &=\sin\left(\dfrac{\left(2\lfloor\sqrt{(n-1)n}\rfloor+1\right)\pi}{2n}\right)+\sum_{k=1}^{n-2}\sin\left(\dfrac{\left(2\lfloor\sqrt{kn}\rfloor+1\right)\pi}{2n}\right)\\ &=\sin\left(\frac{(2n-1)\pi}{2n}\right)+\sum_{j=0}^{n-2}\left(\left\lfloor\frac{(j+1)^2}{n}\right\rfloor-\left\lfloor\frac{j^2}{n}\right\rfloor\right)\sin\left(\frac{(2j+1)\pi}{2n}\right)\\ &=\sin\left(\frac{\pi}{2n}\right)+\sum_{j=0}^{n-2}\Delta\left(\left\lfloor\frac{j^2}{n}\right\rfloor\right)\sin\left(\frac{(2j+1)\pi}{2n}\right)\end{align*}
Using the 1, we get
\begin{align*}S&=\sin\left(\frac{\pi}{2n}\right)+\left[\sin\left(\frac{(2j+1)\pi}{2n}\right)\left\lfloor\frac{j^2}{n}\right\rfloor\right]_{j=0}^{n-1}\\ &{}\quad -\sum_{j=0}^{n-2}\left\lfloor\frac{(j+1)^2}{n}\right\rfloor\left(\sin\left(\frac{(2j+3)\pi}{2n}\right)-\sin\left(\frac{(2j+1)\pi}{2n}\right)\right)\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)\sum_{j=0}^{n-2}\left\lfloor\frac{(j+1)^2}{n}\right\rfloor\cos\left(\frac{(j+1)\pi}{n}\right)\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)\underbrace{\sum_{j=1}^{n-1}\left\lfloor\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)}_{=A}\end{align*}
With sum A, using reverse summand property we get
\begin{align*}A&=\sum_{j=1}^{n-1}\left\lfloor\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)\\ &=\sum_{j=1}^{n-1}\left\lfloor\frac{(n-j)^2}{n}\right\rfloor\cos\left(\frac{(n-j)\pi}{n}\right)\\ &=-\sum_{j=1}^{n-1}\left\lfloor n-2j+\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)\\ &=-A+\sum_{j=1}^{n-1}(2j-n)\cos\left(\frac{j\pi}{n}\right)\\ \Rightarrow A&=\frac{1}{2}\sum_{j=1}^{n-1}(2j-n)\cos\left(\frac{j\pi}{n}\right)\end{align*}
We get
$\displaystyle\cos\left(\frac{j\pi}{n}\right)=\dfrac{1}{2\sin\left(\frac{\pi}{2n}\right)}\left[\sin\left(\frac{(2j+1)\pi}{2n}\right)-\sin\left(\frac{(2j-1)\pi}{2n}\right)\right]=\dfrac{1}{2\sin\left(\frac{\pi}{2n}\right)}\Delta\left[\sin\left(\frac{(2j-1)\pi}{2n}\right)\right]$
continue using 1 :)
$\displaystyle A =\left.\dfrac{(2j-n)}{4\sin\left(\frac{\pi}{2n}\right)}\sin\left(\frac{(2j-1)\pi}{2n}\right)\right|_{j=1}^{n}-\dfrac{1}{4\sin\left(\frac{\pi}{2n}\right)}\sum_{j=1}^{n-1}2\sin\left(\frac{(2j+1)\pi}{2n}\right)$
\begin{align*}A&=\frac{n-1}{2}+\dfrac{1}{4\sin^2\left(\frac{\pi}{2n}\right)}\sum_{j=1}^{n-1}\Delta\left[\cos\left(\frac{j\pi}{n}\right)\right]\\ &=\frac{n-1}{2}+\dfrac{1}{4\sin^2\left(\frac{\pi}{2n}\right)}\cdot\left.\cos\left(\frac{j\pi}{n}\right)\right|_{j=1}^n\\ &=\frac{n-1}{2}-\dfrac{1+\cos\left(\frac{\pi}{n}\right)}{4\sin^2\left(\frac{\pi}{2n}\right)}\\&=\frac{n-1}{2}-\dfrac{2\cos^2\left(\frac{\pi}{2n}\right)}{4\sin^2\left(\frac{\pi}{2n}\right)}\end{align*}
Therefore:
\begin{align*}S&=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)A\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-(n-1)\sin\left(\frac{\pi}{2n}\right)+\dfrac{\cos^2\left(\frac{\pi}{2n}\right)}{\sin\left(\frac{\pi}{2n}\right)}\\ &=\boxed{\displaystyle\cot\left(\frac{\pi}{2n}\right)\cos\left(\frac{\pi}{2n}\right)}\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/404573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Compute eigenvalues and eigenvectors problem I really don't know how solve this problem:
Let $V$ be the space of real functions spanned by $\cos(x)$, $\cos(2x)$ and $\cos(3x)$. Let $T\in\mathcal{L}(V,V)$ con $T(\cos(x)) = 3\cos(x) + 2\cos(2x) - \cos(3x)$, $\;T(\cos(2x)) = 3\cos(2x) + \cos(3x)$ and $T(\cos(3x)) = \cos(3x)$. Find the eigenvalues and eigenvectors of $T$.
Please, I need help
Thanxs in advance.
|
It's clear that
$$T\left(\begin{array}{c}\cos(x)\\\cos(2x)\\\cos(3x)\end{array}\right)\ =\ \underbrace{\left(\begin{array}{ccc}3 & 2 & -1 \\0 & 3 & 1 \\0 & 0 & 1\end{array}\right)}_{A}\left(\begin{array}{c}\cos(x)\\\cos(2x)\\\cos(3x)\end{array}\right)$$
where $A$ have the eigenvalues 1, 3 and 3. Later,
*
*You have that $T(\cos(3x)) = \cos(3x)$, and then $1$ is an eigenvalue of $T$ with eigenvector $v_1 = \cos(3x)$.
*Now, if $v_2 = \alpha\cos(x) + \beta\cos(2x) + \gamma\cos(3x)$ is an eigenvector of $T$ for eigenvector $3$, then
\begin{eqnarray*}
T(v_2) & = & 3v_2,\\
\alpha T(\cos(x)) + \beta T(\cos(2x)) + \gamma T(\cos(3x)) & = & 3\alpha\cos(x) + 3\beta\cos(2x) + 3\gamma\cos(3x),\\
3\alpha\cos(x) + (2\alpha + 3\beta)\cos(2x) - (\alpha - \beta - \gamma)\cos(3x) & = & 3\alpha\cos(x) + 3\beta\cos(2x) + 3\gamma\cos(3x),\\
2\alpha\cos(2x) - (\alpha - \beta + 2\gamma)\cos(3x) & = & 0.
\end{eqnarray*}
Now, $\cos(2x)$ and $\cos(3x)$ are linear independent, then you have
$2\alpha = 0 \Rightarrow \alpha = 0$ and $\alpha - \beta +2\gamma = 0 \Rightarrow \beta = 2\gamma$. Therefore
$$v_2\ =\ \gamma\left[2\cos(2x) + \cos(3x)\right],\quad \gamma \in \mathbb{K}$$
is an eigenvector for 3.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/407352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Proving $\sum_{k=1}^m{k^n}$ is divisible by $\sum_{k=1}^m{k}$ for $ n=2013$ I got an interesting new question, it's about number theory and algebra precalculus. Here is the question:
a positive integer $n$ is called valid if $1^n+2^n+3^n+\dots+m^n$ is divisible by $1+2+3+\dots+m$ for every positive integer $m$.
*
*Prove that 2013 is valid
*Prove that there are infinite positive integers which are not valid
Every little hint, contribution and recommendation would be very helpful. Sorry for my bad english. Thanks before.
|
If $n$ is odd, then modulo $m+1$ we have $2(1^n + 2^n + \ldots + m^n) = (1^n+m^n) + (2^n+(m-1)^n) + \ldots + (m^n+1^n) \\ \equiv (1^n-1^n) + (2^n-2^n) + \ldots + (m^n-m^n) = 0 \pmod {m+1}$.
Also, since $m^n \equiv 0 \pmod m$, $2(1^n + 2^n + \ldots + m^n) \equiv 2(1^n + 2^n + \ldots + (m-1)^n) \equiv 0 \pmod m$
Since $m$ and $m+1$ are coprime, this shows that $2(1^n + \ldots + m^n)$ is a multiple of $m(m+1)$, and since $m(m+1)$ is even, $1^n + \ldots + m^n$ is a multiple of $m(m+1)/2 = 1+2+\ldots+m$.
If $n$ is even then $1+2^n \equiv 2 \pmod 3$, which is not a multiple of $1+2 = 3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/410140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Help in calculating the following integral $\int_0^{2\pi}\! \frac{(1+2\cos x)^n \cos (nx)}{3+2\cos x} \, \mathrm{d}x. $ I was asked to calculate this:
$$\int_0^{2\pi}\! \frac{(1+2\cos x)^n \cos (nx)}{3+2\cos x} \, \mathrm{d}x. $$
My idea was to change the integration limits to $|z|=1$ in the complex plane and to use the residue theorem:
$$\int\limits_{|z|=1}\!\frac{(1+z+z^{-1})^n\frac{1}{2}(z^n+z^{-n})}{3+z+z^{-1}} \,\frac{\mathrm{d}z}{iz} = -\frac{i}{2} \int\limits_{|z|=1}\!\frac{(z^2+z+1)^n(z^{2n}+1)}{z^{2n}(z^2+3z+1)} \,\mathrm{d}z$$
but this requires me to calculate $$\lim_{z \to 0}\frac{d^{2n-1}}{dz^{2n-1}}\left[\frac{(z^2+z+1)^n(z^{2n}+1)}{z^2+3z+1}\right]$$ in order to get the residue at $z=0$. Is there any other way of doing this?
|
$$ \begin{align} \int_{0}^{2 \pi} \frac{(1+2 \cos x)^{n} \cos (nx)}{3 + 2 \cos x} \ dx &= \text{Re} \int_{0}^{2 \pi} \frac{(1+2 \cos x)^{n} e^{inx}}{3 + 2 \cos x} \ dx \\ &= \text{Re} \int_{0}^{2 \pi}\frac{(1+ e^{ix} + e^{-ix})^{n}e^{inx}}{3 + e^{ix} + e^{-ix}} \ dx \\ &= \text{Re} \int_{|z|=1} \frac{(1+z+z^{-1})^{n}z^{n}}{3+z+z^{-1}} \frac{dz}{iz} \\ &= \text{Re} \frac{1}{i} \int_{|z|=1} \frac{(z^{2}+z+1)^{n}}{z^{2}+3z+1} \ dz \\ &= \text{Re} \frac{1}{i} \int_{|z|=1} \frac{(z^{2}+z+1)^{n}}{(z+ \frac{3}{2} - \frac{\sqrt{5}}{2})(z+ \frac{3}{2} + \frac{\sqrt{5}}{2})} \ dz\end{align}$$
Only the pole at $z= - \frac{3}{2} + \frac{\sqrt{5}}{2}$ is inside of the unit circle.
Therefore,
$$ \begin{align} \int_{0}^{2 \pi} \frac{(1+2 \cos x)^{n} \cos (nx)}{3 + 2 \cos x} \ dx &= \text{Re} \frac{1}{i} 2 \pi i \ \text{Res} \Big[\frac{(z^{2}+z+1)^{n}}{z^{2}+3z+1}, - \frac{3}{2} + \frac{\sqrt{5}}{2}\Big] \\ &= 2 \pi \ \text{Re} \lim_{z \to - \frac{3}{2} + \frac{\sqrt{5}}{2}} \frac{(z^{2}+z+1)^{n}}{z+ \frac{3}{2} + \frac{\sqrt{5}}{2}} \\ &= 2 \pi \ \text{Re} \frac{(\frac{7}{2} - \frac{3 \sqrt{5}}{2} - \frac{3}{2} + \frac{\sqrt{5}}{2} +1)^{n}}{\sqrt{5}} \\ &= 2 \pi \frac{(3-\sqrt{5})^{n}}{\sqrt{5}} \end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/410329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
}
|
Compute $\int_{0}^{1}\left[\frac{2}{x}\right]-2\left[\frac{1}{x}\right]dx$ The question is to find
$$\int_{0}^{1}\left(\left[\dfrac{2}{x}\right]-2\left[\dfrac{1}{x}\right]\right)dx,$$
where $[x]$ is the largest integer no greater than $x$, such as $[2.1]=2, \;[2.7]=2,\; [-0.1]=-1.$
Is there any nice method to solve this integral,Thank you everyone.
Some of my thoughts: I have if $a\in(0,1]$, then
$$\int_{0}^{1}\left[\dfrac{a}{x}\right]-a\left[\dfrac{1}{x}\right]dx=a\ln{a}.$$
What about $a>1$,
$$\int_{0}^{1}\left[\dfrac{a}{x}\right]-a\left[\dfrac{1}{x}\right]dx=?$$
and $$\int_{0}^{1}\left(\left[\dfrac{2}{x}\right]-2\left[\dfrac{1}{x}\right]\right)^{2}dx=?$$
becasue I have
$$\int_{0}^{1}\left(\dfrac{1}{x}-\left[\dfrac{1}{x}\right]\right)^2=-1-\gamma+\ln{(2\pi)}$$
|
We have
$$\left\lfloor \dfrac1x \right\rfloor = n \text{ if }x \in \left(\dfrac1{n+1}, \dfrac1n\right]$$
$$\left\lfloor \dfrac2x \right\rfloor = \begin{cases}2n+1 & \text{ if }x \in \left(\dfrac1{n+1}, \dfrac1{n+1/2}\right] \\ 2n & \text{ if }x \in \left(\dfrac1{n+1/2}, \dfrac1{n}\right] \end{cases}$$
We hence have
$$\left\lfloor \dfrac2x \right\rfloor - 2 \left\lfloor \dfrac1x \right\rfloor = \begin{cases} 1 & \text{ if }\left(\dfrac1{n+1}, \dfrac1{n+1/2}\right] \\ 0 & \text{ if }x \in \left(\dfrac1{n+1/2}, \dfrac1{n}\right] \end{cases}$$
Hence,
$$\int_0^1 \left( \left\lfloor \dfrac2x \right\rfloor - 2 \left\lfloor \dfrac1x \right\rfloor\right) dx = \sum_{n=1}^{\infty} \left(\dfrac1{n+1/2} - \dfrac1{n+1}\right) = 2\log2-1$$
One way to evaluate $$\sum_{n=1}^{\infty} \left(\dfrac1{n+1/2} - \dfrac1{n+1}\right)$$ is as follows. We have
$$f(x) = \sum_{n=1}^{\infty} \left(x^{n-1/2} - x^{n} \right) = \sum_{n=1}^{\infty} x^{n-1}\left(\sqrt{x} -x\right) = \dfrac{\sqrt{x}-x}{1-x} = \dfrac{\sqrt{x}}{1+\sqrt{x}}$$
Now we have
$$\int_0^1 f(x) dx = \sum_{n=1}^{\infty} \left(\int_0^1 x^{n-1} dx - \int_0^1 x^{n-1/2} dx\right) = \sum_{n=1}^{\infty} \left(\dfrac1n - \dfrac1{n+1/2}\right)$$
We also have
\begin{align}
\int_0^1 f(x) dx & = \int_0^1 \dfrac{\sqrt{x}dx}{1+\sqrt{x}} = \int_0^1 \dfrac{2t^2dt}{1+t} = 2\int_0^1 \dfrac{dt}{1+t} + 2\int_0^1 \dfrac{(t^2-1)dt}{1+t}\\
& = 2 \log(2) + 2 \int_0^1 (t-1) dt= 2\log(2)-1
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/411612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 1
}
|
probability of who will be selected "In an office there are 3 secretaries, 4 accountants, and 2 receptionists. If a committee of 3 is to be formed, find the probability that one of each will be selected?
Attempted Solution:
First attempt: (3/9)(4/9)(2/9) = 8/243
Second attempt: (3/9)(4/8)(2/7) = 1/21
Don't if either of these are correct or not. Any help would be greatly appreciated to point me in the right direction. Thank you.
|
The standard way to do this is to say there are $\binom{9}{3}$ equally likely ways to choose $3$ people from the $9$. And there are $\binom{3}{1}\binom{4}{1}\binom{2}{1}$ ways to choose one of each kind. For the secretary can be chosen in $\binom{3}{1}$ ways. For each such choice, the accompanying accountant can be chosen in $\binom{4}{1}$ ways. And once this has been done, the receptionist can be chosen in $\binom{2}{1}$ ways.
For the probability, divide "favourables" by "total." We get
$$\frac{\binom{3}{1}\binom{4}{1}\binom{2}{1}}{\binom{9}{3}}.$$
Another way: Imagine choosing the people one at a time. We find the probability of choosing a secretary, then an accountant, then a receptionist.
The probability the first person chosen was an S is $\frac{3}{9}$. Given this happened, the probability the next chosen person was an A is $\frac{4}{8}$. and given this happened, the probability the next person chosen was an S is $\frac{2}{7}$, for a probability of $\frac{3}{9}\cdot\frac{4}{8}\cdot\frac{2}{7}$.
But there are several other ways we could end up with one of each, for example A then S then R. If you calculate, this ends up having the probability $\frac{4}{9}\cdot\frac{3}{8}\cdot\frac{2}{7}$. This is the same number as the one previously obtained.
There are in total $3!=6$ ways we can end up with one of each. so the required probability is
$$6\cdot \frac{3}{9}\cdot\frac{4}{8}\cdot\frac{2}{7}.$$
The first way is (after you get used to it) easier.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/412618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solving the equation $11x^2-6000x-27500 =0$, preferably without the quadratic formula I obtained this form while solving an aptitude question.
$$\frac{3000}{x-50} + \frac{3000}{x+50} = 11$$
I changed it into quadratic equation
$$11x^2 -6000x - 27500 =0$$
but I don't know how to solve it.
I can't find two factor for 303500 that sums to 6000 or when I use formula the numbers become huge...
Without using calculator
how to solve it? is there any other simple way to solve [other method]? [or finding factor] I'm a beginner in math. Please explain your answer for me.
|
$$11x^2 - 6000x - 27500 = 0$$
$a=11$
$b=-6000 = 2^4 \cdot 3 \cdot 5^3$
$c = -27500 = 2^2 \cdot 5^4 \cdot 11$
$ac=-2^2 \cdot 5^4 \cdot 11^2$
Since $ac$ is negative, we will search for integers $u$ and $v$ such that their product is $=2^2 \cdot 5^4 \cdot 11^2$ and their difference is $2^4 \cdot 3 \cdot 5^3 = 6000$. We will sort out exactly where the plusses and minusses go later.
We will try to guess at these numbers by looking at the factors and using a little logic.
If $11 \mid u$ and $11 \mid v$, then $11 \mid u+v$. So both $11's$ must belong to $u$ or $v$ but not both. So we start with this.
\begin{array}{l|c|c|}
& u & v \\
\hline
\text{powers of 2} & & \\
\text{powers of 5} & & \\
\text{powers of 11} & 11^2 & 1\\
\hline
\text{product} & 121\\
\hline
\end{array}
Since $2 \mid u+v$ and $2 \mid uv$, then $u$ and $v$ must each have at least one factor of $2$. So we get this.
\begin{array}{l|c|c|}
& u & v \\
\hline
\text{powers of 2} & 2 & 2\\
\text{powers of 5} & & \\
\text{powers of 11} & 11^2 & 1\\
\hline
\text{product} & 242 & 2\\
\hline
\end{array}
Similarly, $u$ and $v$ must share at least one $5$. So where do we put the other two? Since $6000$ is a pretty big number, we will try putting three of the $5's$ on the same side as the $11's$.
\begin{array}{l|c|c|}
& u & v \\
\hline
\text{powers of 2} & 2 & 2 \\
\text{powers of 5} & 5^3 & 5 \\
\text{powers of 11} & 11^2 & 1 \\
\hline
\text{product} & 30250 & 10\\
\hline
\end{array}
And $u-v$ isn't $6000$.
So let's move one of the $5's$ over.
\begin{array}{l|c|c|}
& u & v \\
\hline
\text{powers of 2} & 2 & 2 \\
\text{powers of 5} & 5^2 & 5^2 \\
\text{powers of 11} & 11^2 & 1 \\
\hline
\text{product} & 6050 & 50\\
\hline
\end{array}
And $u-v=6000$.
Using the $ac$ method, we get
$$\dfrac{(11x-6050)}{11} \dfrac{(11x+50)}{1}$$
$$(x-550)(11x+50)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/413787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
}
|
Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \quad \lambda>\frac{1}{2}$$
Differentiating the above equation with respect to $\lambda$, we obtain an expression involving the Digamma Function $\psi_0(z)$.
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^\lambda}dx = \sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \left(\psi_0(\lambda)-\psi_0 \left( \lambda-\frac{1}{2}\right) \right)$$
Putting $\lambda=2$, we get
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx = -\frac{\pi}{4}+\frac{\pi}{2}\log(2)$$
Question:
But, does anybody know how to evaluate $\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$?
Mathematica gives the values
*
*$\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx = -\frac{G}{6}+\pi \left(-\frac{3}{8}+\frac{1}{8}\log(2)+\frac{1}{3}\log \left(2+\sqrt{3} \right) \right)$
*$\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx = -\frac{\pi}{2}+\frac{\pi \log \left( 6+4\sqrt{2}\right)}{4}$
Here, $G$ denotes the Catalan's Constant.
Initially, my approach was to find closed forms for
$$\int_0^\infty \frac{1}{(1+x^2)^2(1+x^3)^\lambda}dx \ \ , \int_0^\infty \frac{1}{(1+x^2)^2(1+x^4)^\lambda}dx$$
and then differentiate them with respect to $\lambda$ but it didn't prove to be of any help.
Please help me prove these two results.
|
Letting $x\mapsto \frac{1}{x}$ reduces the power of the denominator and changes the integral into
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(1+x^n\right)}{\left(1+x^2\right)^2} d x & =\int_0^{\infty} \frac{x^2 \ln \left(1+x^n\right)}{\left(1+x^2\right)^2} d x-n \int_0^{\infty} \frac{x^2 \ln x}{\left(1+x^2\right)^2} d x \\
& =\int_0^{\infty} \frac{1+x^2-1}{\left(1+x^2\right)^2} \ln \left(1+x^n\right) d x-\frac{n \pi}{4} \\
& =\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x^n\right)}{1+x^2} d x-\frac{n \pi}{8} .
\end{aligned}
$$
Hence
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(1+x^3\right)}{\left(1+x^2\right)^2} d x & =\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x^3\right)}{1+x^2} d x-\frac{3 \pi}{8} \\
& =\frac{1}{2}(-\frac{G}{3}+\frac{\pi}{4} \ln 2 +\frac{2 \pi}{3}\ln(2+\sqrt{3})-\frac{3 \pi}{8} \\
& =-\frac{G}{6}+\pi\left(-\frac{3}{8}+\frac{1}{8} \ln 2+\frac{1}{3} \ln (2+\sqrt{3})\right)
\end{aligned}
$$
The second last line comes from my post.
$$
\begin{aligned}
I_4 & =\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2}-\frac{\pi}{2} \\
& =\frac{\pi \ln (6+4 \sqrt{2})} {4}-\frac{\pi}{2}
\end{aligned}
$$
The second last line comes from my post.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/414642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 6,
"answer_id": 5
}
|
How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't help as it leaves $\frac{1+x-2x^2}{1-4x^2}$ any ideas?
|
When four storey fractions are involved, it's better to do computations separately. Write
$$
N=1-\frac{1-x}{1-2x},\qquad
D=1-2\frac{1-x}{1-2x}
$$
and work on $N$ and $D$:
\begin{align}
N&=1-\frac{1-x}{1-2x}\\
&=\frac{(1-2x)-(1-x)}{1-2x}\\
&=\frac{1-2x-1+x}{1-2x}\\
&=\frac{-x}{1-2x}
\end{align}
and
\begin{align}
D&=1-2\frac{1-x}{1-2x}\\
&=\frac{(1-2x)-2(1-x)}{1-2x}\\
&=\frac{1-2x-2+2x}{1-2x}\\
&=\frac{-1}{1-2x}
\end{align}
Now you know that your fraction is
$$
\frac{N}{D}=N\cdot\frac{1}{D}=\frac{-x}{1-2x}\cdot\frac{1-2x}{-1}
=\frac{-x}{-1}=x
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/415304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
}
|
How to reconcile the identity $\left( e^{i \theta} \right)^{1/2} = e^{i \theta/2}$ with the fact that $a^2 = b$ is solved by $a = \pm\sqrt{b}$ So I worked along the lines of the following:
$$
\left( \cos \left( \theta \right) + i \sin \left( \theta \right) \right)^{\alpha} = \left( e^{i \theta} \right)^{\alpha} = e^{i (\theta \alpha)} = \cos \left( \theta \alpha \right) + i \sin \left( \theta \alpha \right)
$$
with $i$ the imaginary number, $\theta$ real and $\alpha$ real and $\sin , \cos$ the normal trig functions
However if we take $\alpha = \frac{1}{2}$ a fellow member saw that
$$
\left( \cos \theta + i \sin \theta \right)^{1/2} = \left\{ \begin{array}{l} \cos \left( \frac{\theta}{2} \right) + i \sin \left( \frac{\theta}{2} \right) \\ \cos \left( \frac{\theta}{2} + \pi \right) + i \sin \left( \frac{\theta}{2} + \pi \right) \end{array}\right.
$$
which, from my perspective comes down to the fact that $a^2 = b$ can be solved as $a = \sqrt{b} \lor a = -\sqrt{b}$. I feel as though either I'm being very silly (as per usual) or there's something deep going on here that I'm missing
|
Tpofofn's hint is huge:
$$\bullet\;\;\;\left(\cos\frac\theta2+i\sin\frac\theta2\right)^2=\cos^2\frac\theta2-\sin^2\frac\theta2+2i\cos\frac\theta2\sin\frac\theta2=\cos\theta+i\sin\theta$$
$$\bullet\;\;\left(\cos\left(\frac\theta2+\pi\right)+i\sin\left(\frac\theta2+\pi\right)\right)^2=\left(-\cos\frac\theta2-i\sin\frac\theta2\right)^2=\ldots$$
You can complete the exercise above and see that both leftmost expressions above are square roots of the same complex number $\,\cos\theta+i\sin\theta\;\ldots$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/415807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
derive the Maclaurin series by using partial fractions. derive the Maclaurin series for the function
$(x^3+x^2+2x-2)/(x^2-1)$ by using partial fractions and a known Maclaurin series.
question.
how can I use partial fractions in this case?
Is this case the special one?
plz provide me a hint. thanks.
|
Perform polynomial long division
to conclude that
$$\frac{x^{3}+x^{2}+2x-2}{x^{2}-1}=x+1+\frac{3x-1}{x^{2}-1}=x+1+\frac{3x-1}{(x-1)(x+1)}.\tag{1}$$
The theory of partial fractions guarantees that there exist constants $A$ and $B$ such that
$$\frac{3x-1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}.\tag{2}$$
You can find the constants in $(2)$ by several methods, one of them is as follows:
(a) Make e.g. $x=0$ on both sides of $(2)$
$$\frac{-1}{-1}=\frac{A}{-1}+\frac{B}{1}\Leftrightarrow B=A+1\tag{3}.$$
Hence
$$\frac{3x-1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{A+1}{x+1}.\tag{4}$$
(b) Make e.g. $x=1/2$ on both sides of $(4)$
$$\frac{3/2-1}{(1/2-1)(1/2+1)}=\frac{A}{1/2-1}+\frac{A+1}{1/2+1}\Leftrightarrow A=1.\tag{5}$$
Hence $B=2$ and
$$1+x+\frac{3x-1}{(x-1)(x+1)}=1+x+\frac{1}{x-1}+\frac{2}{x+1}.\tag{6}$$
Use the Maclaurin series expansions for $\frac{1}{x-1}$ and $\frac{1}{x+1}$ in $(6)$
$$\frac{1}{x-1}=-\frac{1}{1-x}=-\sum_{n=1}^{\infty }x^{n-1},\quad \frac{1}{x+1}=\frac{1}{1+x}=\sum_{n=1}^{\infty }(-1)^{n-1}x^{x-1},$$
$$\tag{7}$$
both derived from the sum of the geometric series
$$\frac{1}{1-x}=\sum_{n=1}^{\infty }x^{n-1},\qquad (|x|<1).\tag{8}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/416515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Evaluating an integral in physics question $$U_{C} = \frac{1}{C} \int\!\frac{\cos(100\pi t + \pi/4)}{10}\,dt$$
Find $U_{C}$, the answer is $U_{C}=\left(3.2\times 10^{-4}\right)/C\times \cos(100\pi t - \pi/4)$.
Can someone show to to get this answer?
|
First, we make the substitution $u = 100\pi t + \pi/4$, so that $du = 100\pi dt$.
Then, substituting, the integral becomes:
$$U_{c} = \frac{1}{C} \int \cos(u) \cdot \frac{1}{10} \cdot \frac{1}{100\pi}du$$
Integrating, we see that this is:
$$U_{c} = \frac{\frac{1}{1000\pi}}{C}\sin(u) + K$$
or, approximately, with substitution in for $u$:
$$U_{c} = \frac{3.2 \cdot 10^{-4}}{C} \sin(100\pi t + \pi/4) + K$$
Using the fact that $\sin(\theta) = \cos(\theta - \pi/2)$, we see that this is:
$$U_{c} = \frac{3.2 \cdot 10^{-4}}{C} \cos(100\pi t + \pi/4 - \pi/2) + K = \frac{3.2 \cdot 10^{-4}}{C} \cos(100\pi t - \pi/4) + K$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/419616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Jacobi Method and Frobenius Norm Question. I have this linear algebra question concerning the jacobi method and the frobenius norm that I am having a lot of trouble on, I have an exam soon and I would appreciate any help. NOTE: I have read the entire article of wikipedia on jacobi method and frobnenius norm but I have just started the linear algebra topic and there's only so much I can make out of it.
Consider the following system of equations...
$$
\left\{
\begin{array}{c}
2x + y =5\\
x - 3y =6
\end{array}
\right.
$$
1. Compute the Frobenius norm for the Matrix C under the Jacobi iterative method(would you expect this problem to converge)?
2.Starting at (x, y) =
(0, 0) perform 2 iterations of Jacobi Iterative method for this problem.
I dont understand what the Frobenius norm is exactly, so I started at question 2...
First I converted the system of equations to the respective augmented matrix...
$$
\begin{bmatrix}
2 & 1 & 5 \\
1 & -3 & 6 \\
\end{bmatrix}
$$
From here I can see that...
$$
\left\{
\begin{array}{c}
x^{new} = (-y^{old} + 5)/2\\
y^{new} = (6 - x^{old})/-3
\end{array}
\right.
$$
So I set up the array...
$$
\begin{array}{c|lcr}
Iteration & \text{x^{old}} & \text{y^{old}} & \text{x^{new}} & \text{y^{new}}\\
\hline
1 & 0 & 0 & 2.5 & -2 \\
2 & 2.5 & -2 & 3.5 & -7/6 \\
\end{array}
$$
And that is my answer to number 2, any help with how to do number 1 would be greatly appreciated, thank you for your time.
|
Hint: The Frobenius norm of an $m \times n$ matrix $A$ is defined as the square root of the sum of the absolute squares of its elements.
Example: Consider matrix
$$A = \left(
\begin{array}{ccc}
~~~~1 & -2 & ~~~~~~3 \\
-4 & ~~5 & ~-6 \\
~~~7 & -8 & ~~~~~~9 \\
\end{array}
\right)$$
$\lVert A \rVert _{F} = \sqrt (|1|^2+|-2|^2+|3|^2+|-4|^2+|5|^2+|-6|^2+|7|^2+|-9|^2 + |-8|^2) = \sqrt(285) = 16.8819$
For convergence of Jacobi iteration method you need to find iteration matrix $P = -D^{-1}(L+U)$. Note that the Jacobi iterative scheme will converge if $\lVert P\rVert_{F} $ is strictly less than $1$, where $F$ stands for the Frobenius norm. There may be some other matrix norm (such as the $1$-norm ) that is strictly less than $1$, in which case convergence is still guaranteed. In any case, however, the condition $\lVert P\rVert <1 $ is only a sufficient condition for convergence, not a
necessary one.
For the given example
$D = \left(
\begin{array}{ccc}
~~1 & ~0 & ~0 \\
~~0 & ~5 & ~0 \\
~~0 & ~0 & ~9 \\
\end{array}
\right)$
$L = \left(
\begin{array}{ccc}
~~~0 & ~~0 & ~0 \\
-4 & ~~0 & ~0 \\
~~~~7 & -8 & ~0 \\
\end{array}
\right)$
$U = \left(
\begin{array}{ccc}
~~0 & -2 &~~3 \\
~~0 & ~~~0 & -6 \\
~~0 & ~~~0 & ~~~0 \\
\end{array}
\right)$
Added: Consider to solve $3\times 3$ size system of linear equation $Ax = b$, where coefficient matrix
$A = \left(
\begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
a_{21}& a_{22} & a_{23} \\
a_{31}& a_{32} & a_{32} \\
\end{array}
\right)$
Assume coefficient matrix $A$ has no zeros on its main diagonal i.e. $a_{11}$, $a_{22}$, $a_{23}$ are non zeros, then
$D = \left(
\begin{array}{ccc}
~~\frac{1}{a_{11}} & 0 & 0 \\
0 & \frac{1}{a_{22}} & ~0 \\
~~0 & 0 & \frac{1}{a_{33}}\\
\end{array}
\right)$
$L = \left(
\begin{array}{ccc}
0 & 0 & 0 \\
a_{21} & 0 & ~0 \\
a_{31}& a_{32} & 0\\
\end{array}
\right)$
$U = \left(
\begin{array}{ccc}
0 & a_{12} & a_{13} \\
0 & 0 & a_{23} \\
0 & 0 & 0\\
\end{array}
\right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/419930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
Let
$$\begin{align*}x &= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} + \cdots\\
y &= \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots\\
z &= \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots \end{align*}$$
so we have
$$x = y + z.$$
However, $x = 2\cdot z$, so $y$ = $z$ or
$$\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots$$
This looks ok if I interpret it as
$$\frac{1}{1} = \left (\frac{1}{2} - \frac{1}{3} \right ) + \left (\frac{1}{4} - \frac{1}{5} \right ) + \left (\frac{1}{6} - \frac{1}{7} \right ) + \cdots + \left (\frac{1}{2n} - \frac{1}{2n+1} \right ) + \cdots$$
However, it's a bit weird if I write it as
$$\left (\frac{1}{1} - \frac{1}{2} \right ) + \left (\frac{1}{3} - \frac{1}{4} \right ) + \left (\frac{1}{5} - \frac{1}{6} \right ) + \cdots + \left (\frac{1}{2n-1} - \frac{1}{2n} \right ) + \cdots = 0.$$
How can a sum of positive numbers equal $0$?
|
You are manipulating divergent series. When you do the substraction, you can in fact obtain any limit you want by changing the order in which you are adding or subtracting things.
For example,
$(1 - \frac 12 - \frac 14 - \frac 18 - \ldots) + (\frac 13 - \frac 16 - \frac 1{12} - \frac 1{24}- \ldots) + (\frac 15 - \frac 1{10} - \frac 1{20} - \frac 1{40} - \ldots) + \ldots = 0 + 0 + 0 + \ldots = 0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/420047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
}
|
$\frac{c}{1+c}\leq\frac{a}{1+a}+\frac{b}{1+b}$ for $0\leq c\leq a+b$ When proving that if $d$ is a metric, then $d'(x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is also a metric, I have to prove the inequality:
$$\dfrac{c}{1+c}\leq\frac{a}{1+a}+\frac{b}{1+b}$$ for $0\leq c\leq a+b$. This is obvious by expanding, but is there a nicer way to see why it is true?
|
$$\frac{a}{1+a}+\frac{b}{1+b}\ge\frac{a}{1+a+b}+\frac{b}{1+b+a}=\frac{a+b}{1+a+b}=\frac{1}{1+\frac{1}{a+b}}\ge\frac{1}{1+\frac1c}=\frac{c}{1+c}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/421099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
multiplication in GF(256) (AES algorithm) I'm trying to understand the AES algorithm in order to implement this (on my own) in Java code.
In the algorithm all byte values will be presented as the concatenation of its individual bit values (0 or 1) between braces with the most significant bit first.
So bytes are interpreted as finite field elements using a polynomial representation:
$$b_7x^7+b_6x^6+b_5x^5+b_4x^4+b_3x^3+b_2x^2+b_1x+b_0$$
Example: ${01100011} \equiv x^6 + x^5 + x +1$
These finite field elements can easily be added with the XOR operator.
But multiplication is a little bit more complex, and I don't understand it.
In the polynomial representation, multiplication in GF(256) corresponds with the multiplication of polynomials modulo an irreducible polynomial of degree 8
This irreducible polynomial is: $m(x)=x^8 +x^4 +x^3 +x+1$
Here is my example:
$$01010111 * 00010011$$
$$01010111\equiv (x^6+x^4+x^2+x+1)$$
$$00010011\equiv(x^4+x+1)$$
$$(x^6+x^4+x^2+x+1) * (x^4+x+1) = x^{10}+x^8+x^7+x^3$$
$$(x^{10}+x^8+x^7+x^3) \mod m(x)$$
$$(x^{10}+x^8+x^7+x^3) \mod (x^8 +x^4 +x^3 +x+1) = x^2 + 1$$
$$x^{10} x^3$$
$$x^8+x^7+x^6+x^5+x^2$$
$$x^8$$
$$x^7+x^6+x^5+x^4+x^3+x^2+x+1$$
So here $x^7+x^6+x^5+x^4+x^3+x^2+x+1$ is my solution, that leads to a binary representation: $11111111$
But in the literature the correct solution is: $11111110$
I would be happy if you could tell me where my mistake is.
|
$$(x^6+x^4+x^2+x+1) * (x^4+x+1) = x^{10}+x^8+x^7+x^3$$
should make you suspicious. What happened to the cross-term $1*1$?
I make that product to evaluate to $x^{10}+x^8+x^7+x^3+1$, and the difference of $+1$ should carry through the remaining computation to give the difference you sought.
The modulo operation is equivalent to assuming $x^8 = x^4+x^3+x+1$, which gives $x^{10} = x^6+x^5+x^3+x^2$ so $$\begin{eqnarray*}x^{10}+x^8+x^7+x^3+1 & = & (x^6+x^5+x^3+x^2) + (x^4+x^3+x+1) + x^7+x^3+1\\
& = & x^7+x^6+x^5+x^4+x^3+x^2+x\end{eqnarray*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/422613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Indefinite integration : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$ Problem :
Solve : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$
I tried :
$\frac{1-x^2}{\sqrt{(1-x^2)^3}} + \frac{x}{\sqrt{(1-x^2)^3}}$
But it's not working....Please guide how to proceed
|
Hint:when $|x|\lt 1$
$$\int(\frac{1-x^2}{\sqrt{(1-x^2)^3}} + \frac{x}{\sqrt{(1-x^2)^3}})dx=\int\frac{dx}{\sqrt{(1-x^2)}} + \frac{-1}{2}\int\frac{-2xdx}{\sqrt{(1-x^2)^3}}$$ take $$u=1-x^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/423286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
}
|
Evaluating the definite integral $\int^{\pi/2}_0 \frac{x+\sin x}{1+\cos x}\,\mathrm dx$ Problem :
$$\int^{\pi/2}_0 \frac{x+\sin x}{1+\cos x}\,\mathrm dx \tag{i}$$
My approach :
$$\frac{x+\sin x}{1+\cos x}\,\mathrm dx= \left ( \frac{x}{2\cos^2(x/2)} + \tan(x/2) \right )\,\mathrm dx$$
Therefore, (i) will become :
$$\int^{\pi/2}_0\left (\frac{x }{2\cos^2(x/2)} + \frac{1}{2}\tan(x/2) \right )\,\mathrm dx = \int^{\pi/2}_0 \frac{1}{2}x \sec^2(x/2)\,\mathrm dx + \int^{\pi/2}_0 \frac{1}{2}\tan(x/2)\,\mathrm dx$$
Please guide how to proceed. Thanks.
|
Since $\displaystyle\int\dfrac{x+\sin x}{1+\cos x}dx$ has close-form, I don't think this question is really difficult.
Since
$$
\begin{align}
\int\dfrac{x+\sin x}{1+\cos x}\text{d}x & = \int\dfrac{x}{1+\cos x}dx+\int\dfrac{\sin x}{1+\cos x}dx \\
& = \int\dfrac{x}{2\cos^2\dfrac{x}{2}}dx+\int\tan\dfrac{x}{2}dx \\
& = \int\dfrac{x}{2}\sec^2\dfrac{x}{2}dx+\int\tan\dfrac{x}{2}dx \\
& = \int x~d\left(\tan\dfrac{x}{2}\right)+\int\tan\dfrac{x}{2}dx \\
& = x\tan\dfrac{x}{2}-\int\tan\dfrac{x}{2}dx+\int\tan\dfrac{x}{2}dx \\
& =x\tan\dfrac{x}{2}+C
\end{align}
$$
we have that
$$
\int_0^{\frac{\pi}{2}}\dfrac{x+\sin x}{1+\cos x}dx=\left[x\tan\dfrac{x}{2}\right]_0^{\frac{\pi}{2}}=\dfrac{\pi}{2}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/423757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Given that $ x + \frac 1 x = r $ what is the value of: $ x^3 + \frac 1 {x^2}$ in terms of $r$? Given that $$ x + \cfrac 1 x = r $$
what is the value of: $$ x^3 + \cfrac 1 {x^2}$$ in terms of $r$?
NOTE: it is $\cfrac 1 {x^2}$ and not $ \cfrac 1 {x^3} $
Where I reached so far:
$$ \Big(x^3 + \cfrac 1 {x^2}\Big) + \cfrac 1 x \cdot\Big(x^3 + \cfrac 1 {x^2}\Big) = r^3 - r^2 -3r - 2 $$
Any hints??
|
The answer could be found by direct calculation:
$$ x = \frac{r \pm \sqrt{r^2-4}}{2}$$
$$x^3+\frac{1}{x^2} = -\frac{(- r^2 + r + 1)(r + \sqrt{r^2 - 4} + 2)}{2} \tag{$+$}$$
$$x^3+\frac{1}{x^2} = \frac{4}{(r - \sqrt{r^2 - 4})^2 + (r - \sqrt{r^2 - 4})^{\frac{3}{8}}}\tag{$-$}$$
Edit1 Please note that the domain of $r$ is
$$r\in \{\mathbb{R}-(-2,+2)\}$$
for example, if $r=\cfrac{5}{2}$, then $x=2,\cfrac{1}{2}$ and
$$x^3+\frac{1}{x^2}=\frac{33}{4}, \frac{33}{8}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/424471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Evaluate $\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx$ I am trying to find a closed form for
$$\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx = 0.094561677526995723016 \cdots$$
It seems that the answer is
$$\frac{\pi^2}{12}\left( 1-\sqrt{3}\right)+\log(2) \log \left(1+\sqrt{3} \right)$$
Mathematica is unable to give a closed form for the indefinite integral.
How can we prove this result? Please help me.
EDIT
Apart from this result, the following equalities are also known to exist:
$$\begin{align*}
\int_0^1 \frac{\log \left( 1+x^{4+\sqrt{15}}\right)}{1+x}\mathrm dx &=\frac{\pi^2}{12} \left( 2-\sqrt{15}\right)+\log \left( \frac{1+\sqrt{5}}{2}\right)\log \left(2+\sqrt{3} \right) \\ &\quad +\log(2)\log\left( \sqrt{3}+\sqrt{5}\right)
\\ \int_0^1 \frac{\log \left( 1+x^{6+\sqrt{35}}\right)}{1+x}\mathrm dx &= \frac{\pi^2}{12} \left( 3-\sqrt{35}\right)+\log \left(\frac{1+\sqrt{5}}{2} \right)\log \left(8+3\sqrt{7} \right) \\
&\quad +\log(2) \log \left( \sqrt{5}+\sqrt{7}\right)
\end{align*}$$
Please take a look here.
|
I have found an interesting reference related to this integral. It may be of interest to many users:
A restricted Epstein zeta function
and the evaluation of some definite integrals by Habib Muzaffar and Kenneth S. Williams
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/426325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "169",
"answer_count": 3,
"answer_id": 0
}
|
How to derive compositions of trigonometric and inverse trigonometric functions? To prove:
$$\begin{align}
\sin({\arccos{x}})&=\sqrt{1-x^2}\\
\cos{\arcsin{x}}&=\sqrt{1-x^2}\\
\sin{\arctan{x}}&=\frac{x}{\sqrt{1+x^2}}\\
\cos{\arctan{x}}&=\frac{1}{\sqrt{1+x^2}}\\
\tan{\arcsin{x}}&=\frac{x}{\sqrt{1-x^2}}\\
\tan{\arccos{x}}&=\frac{\sqrt{1-x^2}}{x}\\
\cot{\arcsin{x}}&=\frac{\sqrt{1-x^2}}{x}\\
\cot{\arccos{x}}&=\frac{x}{\sqrt{1-x^2}}
\end{align}$$
|
I'll be ignoring domains and possible roots of negative numbers. (If you let $\mbox{$x\in \textbf{]}0,\pi/2[$}$ everything works fine).
Given $f\circ g$, the trick is too relate $f$ with $g^{-1}$.
I did some. You should be able to handle the remaining ones.
$\bullet \sin (\arccos (x))=\sqrt {1-(\cos (\arccos (x) ))^2}=\sqrt {1-x^2}$
$\bullet \sin (\arctan (x))=\dfrac{\tan (\arctan (x))}{\sqrt {1+(\tan (\arctan (x)))^2}}=\dfrac{x}{\sqrt {1+x^2}}$
For this one I used
$$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies 1+(\tan(x))^2=(\sec(x))^2\\
&\implies 1+(\tan (x))^2=\dfrac{1}{1-(\sin (x))^2}\\
&\implies 1-(\sin(x))^2=\dfrac{1}{1+(\tan(x))^2}\\
&\implies \sin (x)=\dfrac{\tan(x)}{\sqrt{1+(\tan(x))^2}}\end{align}$$
$\bullet \tan (\arcsin (x))=\dfrac{\sin (\arcsin (x))}{\sqrt{1-(\sin(\arcsin (x)))^2}}=\dfrac{x}{\sqrt{1-x^2}}$
For this one I used
$$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies 1+(\tan(x))^2=(\sec(x))^2\\
&\implies 1+(\tan (x))^2=\dfrac{1}{1-(\sin (x))^2}\\
&\implies \tan(x)=\dfrac{\sin(x)}{\sqrt{1-(\sin(x))^2}}\end{align}$$
$\bullet \cot (\arcsin(x))=\dfrac{\sqrt{1-(\sin (\arcsin(x)))^2}}{\sin (\arcsin(x))}=\dfrac{\sqrt{1-x^2}}{x}$
For this one I used
$$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies (\cot (x))^2+1=(\csc(x))^2\\
&\implies (\cot (x))^2=\dfrac{1-(\sin(x))^2}{(\sin(x))^2}\\
&\implies \cot(x)=\dfrac{\sqrt{1-(\sin(x))^2}}{\sin (x)}\end{align}$$
$\bullet \cot (\arccos(x))=\dfrac{\cos (\arccos(x))}{\sqrt{1-(\cos (\arccos(x)))^2}}=\dfrac{x}{\sqrt{1-x^2}}$.
For this one I used
$$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies (\cot (x))^2+1=(\csc(x))^2\\
&\implies (\cot (x))^2=\dfrac{1-(\sin(x))^2}{(\sin(x))^2}\\
&\implies (\cot(x))^2=\dfrac{(\cos(x))^2}{1-(\cos(x))^2}\\
&\implies \cot(x)=\dfrac{\cos(x)}{\sqrt{1-(\cos(x))^2}}\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/426399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
}
|
Trigonometric near-identity The parametrized curve
$$
\left( \sec\theta+\csc\theta,\ 2\sqrt{2}\csc(2\theta) \right), \qquad \frac{10}{100} \le\theta\le\frac{142}{100}
$$
looks to the naked eye like a straight line. The $y$-intercept is not $0$ and the slope is a number that I haven't tried to make sense out of (yet). (The factor $2\sqrt{2}$ was chosen only to make the minimum values of the two coordinates equal to each other.) The best-fitting straight line gives all residuals less than $0.08$, quite small!
Why?
|
Call $K=\csc\left(\theta\right)+\sec\left(\theta\right)$ and $W=a/\sin\left(2\theta\right)$ for some $a\gt0$ and $\theta\in\left]0,\pi/2\right[$ instead. You are interested in why $K$ and $W$ almost have a linear relationship. First, we know that $$K^2=\frac{1}{\sin^2\left(\theta\right)}+\frac{1}{\cos^2\left(\theta\right)}+\frac{2}{\sin\left(\theta\right)\cos\left(\theta\right)}\\=\frac{4a^2}{a^2\cdot 4\sin^2\left(\theta\right)\cos^2\left(\theta\right)}+\frac{4a}{a\cdot 2\sin\left(\theta\right)\cos\left(\theta\right)}=\frac{4W^2}{a^2}+\frac{4W}{a}\\\Longrightarrow K=\frac{2W}{a}\left(1+\frac{a}{W}\right)^{1/2}\\ \Longrightarrow \csc\left(\theta\right)+\sec\left(\theta\right)=\frac{2\sqrt{1+\sin(2\theta)}}{\sin\left(2\theta\right)}.$$ This is an exact identity. In the given interval, $\displaystyle \left|\sin\left(2\theta\right)\right| =\left|\frac{a}{W}\right|\lt 1$. We can invoke the binomial theorem: $$K=\frac{2}{\sin\left(2\theta\right)}\left[1+\frac{\sin\left(2\theta\right)}{2}-\frac{\sin^2\left(2\theta\right)}{8}+\text{higher order terms}\right]\approx \frac{2}{\sin\left(2\theta\right)}+1$$ as the other terms are going to zero. As said, this is almost a linear relationship. How good is it? Not much.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/427195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
}
|
Show that $ \mathbf u^2 \mathbf v^2 = (\mathbf u \cdot \mathbf v)^2 - (\mathbf u \wedge \mathbf v)^2 $ where $ \mathbf u $ and $ \mathbf v $ are vectors. From Linear and Geometric Algebra by Alan Macdonald.
|
You can use
$$
\begin{align}
(u\cdot v)^2 - (u\wedge v)^2
& = \frac{1}{4} \left( (uv+vu)^2 - (uv-vu)^2\right) \\
& = \frac{1}{4} \left(uvuv + uvvu + vuuv + vuvu - uvuv + uvvu + vuuv - vuvu \right) \\
& = \frac{1}{2} \left( uvvu + vuuv\right) \\
& = \frac{1}{2} \left( u^2v^2+u^2v^2\right) \\
& = u^2v^2
\end{align}
$$
The third line to the fourth line is true because
$$uvvu = uv^2u = uuv^2 = u^2v^2$$
where we can re-order terms because $u^2$, $v^2$ are scalars.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/427433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Show that $e^x \geq (3/2) x^2$ for all non-negative $x$ I am attempting to solve a two-part problem, posed in Buck's Advanced Calculus on page 153. It asks "Show that $e^x \geq \frac{3}{2}x^2$ $\forall x\geq 0$. Can $3/2$ be replaced by a larger constant?"
This is after the section regarding Taylor polynomials, so I have been attempting to leverage the Taylor expansion for $e^x$ at $0$. $e^x \geq 1+x+\frac{x^2}{2}$, and by quadratic formula, we have $1+x+\frac{x^2}{2}\geq \frac{3}{2}x^2$ for $x\in [0, \frac{1+\sqrt{5}}{2}]$.
Now also $e^x\geq 1+x+\frac{x^2}{2}+\frac{x^3}{6}$. We know there exists a $c\in \mathbb{R}^{\geq 0}$ such that for all $x\geq c$, we have $1+x+\frac{x^2}{2}+\frac{x^3}{6}\geq \frac{3}{2}x^2$. I want to find this point without messing with the cubic formula, etc. I think I am missing a simpler way. Any ideas?
|
Use AM-GM (i. e. $a+b \ge 2\sqrt{ab})$:
$$ x+x^3/6 \ge 2\sqrt{x^4/6}=x^2\sqrt{2/3}$$
$$1+x^4/24 \ge 2\sqrt{x^4/24}=x^2\sqrt{1/6}$$
Thus for all $x \ge 0$:
$$e^x \ge 1+x+x^2/2+x^3/6+x^4/24\ge x^2(1/2+ \sqrt{2/3}+\sqrt{1/6}) \ge 3/2 x^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/427500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
}
|
prove that $\cos x,\cos y,\cos z$ don't make strictly decreasing arithmetic progression let $x,y,z\in R$,and such that $$\sin y-\sin x=\sin z-\sin y\ge 0 $$ show that:
$$\cos x,\cos y,\cos z$$ don't make strictly decreasing arithmetic progression
my idea:
we have $$2\sin y=\sin x +\sin z\cdots\cdots\tag 1$$
and assume that,there exist $x,y,z$ such that
$$2\cos y=\cos x+\cos z\cdots\cdots \tag2$$
and $(1)^2+(2)^2$,we have
$$4=2+2(\sin x\sin z+\cos x\cos z)=2+2\cos(x-z)$$
then
$$\cos(x-z)=1\Longrightarrow x=z+k\pi,k\in Z$$
so
$$\cos x=(-1)^k\cos z,\sin x=(-1)^k\sin z$$
Then ?
|
We have $\sin y = \sin x + p$ and $\sin z = \sin x+2p$, where $p$ is positive.
Suppose that $\cos y = \cos x - r$ and $\cos z = \cos x - 2r$ for positive $r$.
Then $\cos^2 y = 1-(\sin x+p)^2 = (\cos x - r)^2$ and $\cos^2 z = 1-(\sin x+2p)^2 = (\cos x - 2r)^2$.
So $\cos^2y = 1-\sin^2 x-p^2-2p\sin x = \cos^2 x - 2r\cos x + r^2$ and $\cos^2 z = 1-\sin^2 x - 4p^2 - 4p\sin x = \cos^2 x -4r\cos x + 4r^2$.
Then $-p^2-2p\sin x = -2r\cos x + r^2$ and $-4p^2 - 4p\sin x=-4r\cos x+4 r^2$.
That last gets you $p^2+p\sin x = r\cos x+r^2$.
I have to run now.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/428748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
}
|
Showing that $\lim\limits_{n\to\infty}x_n$ exists, where $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$ Let $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$
a) Show that $x_{n} < x_{n+1}$
b) Show that $x_{n+1}^{2} \leq 1+ \sqrt{2} x_{n}$
Hint : Square $x_{n+1}$ and factor a 2 out of the square root
c) Hence Show that $x_{n}$ is bounded above by 2. Deduce that $\lim\limits_{n\to \infty} x_{n}$ exists.
Any help? I don't know where to start.
|
Hints: Induction on
$$\bullet\;\;x_n<x_{n+1}\iff 1+\sqrt{2+\sqrt{3\ldots+\sqrt n}}<1+\sqrt{2+\sqrt{3+\ldots+\sqrt{n+\sqrt{n+1}}}}\iff$$
$$2+\sqrt{3+\ldots+\sqrt n}<2+\sqrt{3+\ldots\sqrt{n+1}}\iff\ldots$$
$$\bullet\bullet\;x_{n+1}^2=1+\sqrt{2+\sqrt{3+\ldots+\sqrt{n+1}}}\le 1+\sqrt2\left(\sqrt{1+\sqrt{2+\ldots+\sqrt n}}\right)=1+\sqrt2\,x_n$$
$$\iff\left(\sqrt{2+\sqrt{3+\ldots+\sqrt{n+1}}}\right)\le\sqrt{2+2\sqrt{2+\ldots+\sqrt n}}\iff\ldots$$
For (c) you're already done with (a)-(b) since then you have a monotone ascending sequence bounded from above, so the sequence's limit equals its supremum...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/428841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Seeking a combinatorial proof of the identity $1+2\cdot 2^1+3\cdot 2^2+\cdots+n\cdot 2^{n-1}=(n-1)2^n+1$ I would appreciate if somebody could help me with the following problem
Q: show that combinatoric identity (using by combinatorial proof)
$$1+2\cdot 2^1+3\cdot 2^2+\cdots+n\cdot 2^{n-1}=(n-1)2^n+1$$
|
Combinatoric interpretation.
Sub-task:
There is a ordered sequence (array, vector, chain, cortege, ...) of $k$ balls.
1 ball $-$ red, other balls $-$ black or white.
Let $M(k)$ $-$ number of such arrays.
$M(k) = k \cdot 2^{k-1}$.
Example for $k=3$:
$\Huge{\color{red}\bullet} \circ\circ$,
$\Huge{\color{red}\bullet} \circ\bullet$,
$\Huge{\color{red}\bullet} \bullet\circ$,
$\Huge{\color{red}\bullet} \bullet\bullet$;
$\Huge{\circ\color{red}\bullet} \circ$,
$\Huge{\circ\color{red}\bullet} \bullet$,
$\Huge{\bullet\color{red}\bullet} \circ$,
$\Huge{\bullet\color{red}\bullet} \bullet$;
$\Huge{\circ\circ\color{red}\bullet}$,
$\Huge{\circ\bullet\color{red}\bullet}$,
$\Huge{\bullet\circ\color{red}\bullet}$,
$\Huge{\bullet\bullet\color{red}\bullet}$.
Task:
To find $S(n)$: number of all possible arrays (with 1 red and other black/white balls) with length $k\leqslant n$.
$S(n) = M(1)+M(2)+\ldots+M(n)$.
How to prove:
Let $L(k)$ $-$ number of arrays, where red ball is on the $k$-th place:
$L(1) = 1+2+4+\cdots+2^{n-1} = 2^n-1$;
$L(2) = 2+4+\cdots+2^{n-1} = 2^n-2^1$;
$\ldots$
$L(k) = 2^{k-1}+2^k+\cdots+2^{n-1} = 2^n-2^{k-1}$;
$\ldots$
$L(n-1) = 2^{n-2}+2^{n-1} = 2^n-2^{n-2}$;
$L(n) = 2^{n-1} = 2^n-2^{n-1}$.
So,
$$
S(n) = \sum_{k=1}^n L(k) = \sum_{k=1}^{n} (2^n - 2^{k-1}) = n2^n - (2^n-1) = (n-1)2^n+1.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/429084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
}
|
Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ Problem statement:
Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $.
, $n\in \mathbb{N}$
My progress
LHS is equivalent to $\frac{(2n-1)!}{(2n)!}=\frac{(2n-1)(2n-2)(2n-3)\cdot ....}{(2n)(2n-1)(2n-2)\cdot ....}=\frac{1}{2n}$ So we can rewrite our inequality as:
$\frac{1}{2n}\leq \frac{1}{\sqrt{2n+1}}$ Let's use induction:
For $n=1$ it is obviously true. Assume $n=k$ is correct and show that $n=k+1$ holds.
$\frac{1}{2k+2}\leq \frac{1}{\sqrt{2k+3}}\Leftrightarrow 2k+2\geq\sqrt{2k+3}\Leftrightarrow 4(k+\frac{3}{4})^2-\frac{5}{4}$ after squaring and completing the square. And this does not hold for all $n$
About convergence: Is it not enough to check that $\lim_{n \to \infty}\frac{1}{2n}=\infty$ and conclude that it does not converge?
|
Suppose that $P(n)$ which is $\frac{1}{2}\frac{3}{4} \dots \frac{2n-1}{2n} \lt \frac{1}{\sqrt{3n+1}}$. Now $P(n + 1)$ would be $\frac{1}{2}\frac{3}{4} \dots \frac{2n-1}{2n}\frac{2n+1}{2n+2} \overset{?}{\lt} \frac{1}{\sqrt{3n+4}} $. Using the induction hypothesis it would be enough to prove that $\frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2} \lt \frac{1}{\sqrt{3n+4}} \leftrightarrow (2n+1)^2(3n+4) \lt (2n+2)^2(3n+1) \leftrightarrow 12n^3+28n^2+19n+4 $$\lt 12n^3+28n^2+20n+4$ which is true.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/431234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
}
|
On the period of the decimal representation of $n$ when $\gcd(n, 10) \neq 1$ Suppose that $n = 2^a5^bm$, where $n > m > 1$ are integers, with $\gcd(m, 10) = 1$, and $a, b$ are non-negative integers.
How does one show that the lengths of the periods of the decimal expansions of $1/n$ and $1/m$ are the same?
I know that if the length of the period of the decimal expansion of $m$ is $r > 0$, then $10^r\equiv 1\;(\!\!\!\!\mod m)\;$, and furthermore, that $r$ is the smallest positive integer for which this congruence holds...
|
[Originally, some of the content below was included as a sort of appendix to the original question, but once I managed to complete the proof (I think), I decided to post it as an answer.]
If the length of the period of the decimal expansion of $m$ is $r > 0$, then $10^r\equiv 1\;(\!\!\!\!\mod m)\;$, and furthermore, that $r$ is the smallest positive integer for which this congruence holds.
Therefore, for some positive integer $s$, we have $s m = 10^r - 1$. Since $m > 1$, we have that $0 < s < 10^r$, so the decimal representation of $s$ contains at most $r$ digits, and
$$\frac{1}{m} = \frac{s}{10^r} \sum_{k=0}^\infty 10^{-kr} = \frac{s\sigma}{10^r},$$
...where, for the last expression, I've defined $\sigma = \sum_{k=0}^\infty 10^{-kr}$.
Now, define $t = 2^{\max(0,b-a)}5^{\max(0,a-b)}$, and $u = \max(a, b)$. Then we can write
$$\frac{1}{n} = \frac{1}{2^a5^bm} = \frac{t}{10^u}\cdot\frac{1}{m} = \frac{ts\sigma}{10^{r + u}}$$
Of course, the claim is trivially true if $a = b$, so let's assume that $a \neq b$. Hence $t > 1$ and $u > 0$.
Now, using the standard identity
$$
\sigma = \sum_{k=0}^\infty 10^{-kr} = \frac{1}{1-10^{-r}} = \frac{10^r}{10^r - 1},
$$
we define $v = 10^r - 1 = 10^r/\sigma$, and let $w$ and $x$ be such that $0 \leq x < v$ and
$$
ts = vw + x.
$$
Then,
$$
\frac{1}{n} = \frac{ts\sigma}{10^{r+u}} = \frac{vw\sigma}{10^{r+u}} + \frac{x\sigma}{10^{r+u}}
$$
But, by definition of $v$, $(v \sigma)/10^r= 1$. Hence
$$
\frac{1}{n} = \frac{w}{10^u} + \frac{x}{10^{r+u}} \sum_{k=0}^\infty 10^{-kr}
=\frac{1}{10^u} \left( w + \frac{x}{10^r} \sum_{k=0}^\infty 10^{-kr} \right).
$$
All that remains is to show that $x > 0$, or, IOW, that $v$ does not divide $ts$. Since $t$ is either a power of $2$ or a power of $5$, whereas $v = 10^r - 1$, it follows that $\gcd(t, v) = 1$. Therefore, $v|ts$ only if $v|s$. But $v = sm$, and $m > 1$, so $0< s < v$, and therefore $v\nmid s$. So we can conclude that $v\nmid ts$, and therefore, $x > 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/432518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Find the minimum of this expression This is a problem in my exam and I can't find the solution using elementary inequality knowledge. Can anyone here help me solve this. Thanks
$a,b,c $ are positive real numbers which satisfy $(a+c)(b+c) = 4c^{2}$. Find the minimum of this expression:
$$P = \frac{32a^{3}}{(b+3c)^{3}} + \frac{32b^3}{(a+3c)^{3}} - \frac{\sqrt{a^{2} + b^{2}}}{c}$$
Thanks so much.
|
We begin with the observation that the change $a=xc, b=yc$ reduces the problem under consideration to the two-dimensional optimization problem
$$\min \frac {x^3} {(y+3)^3} + \frac {y^3} {(x+3)^3}-\sqrt{x^2+y^2}$$
under the constraints $$(x+1)(y+1)=4, x \ge 0, y \ge 0 . $$
Solving it with Mathematica by
Minimize[ 32 x^3/(y + 3)^3 + 32 y^3/(x + 3)^3 - Sqrt[x^2 + y^2],
(x + 1)*(y + 1) == 4 && x >= 0 && y >= 0, {x, y}]
we obtain:
{1 - \sqrt{2}, {x -> 1, y -> 1}}
i.e. $\{a=b,b=c\}.$ One may play with Lagrange multipliers to this end.
Addition. For each nonnegative $x$ and for each nonnegative $y$ the inequality
$$ \frac {x^3} {(y+3)^3} + \frac {y^3} {(x+3)^3}-\sqrt{x^2+y^2} \ge 2\sqrt{\frac {x^3y^3} {(x+3)^3(y+3)^3}}-\sqrt{x^2+y^2}$$ holds.
The equality takes place iff $\frac {x^3} {(y+3)^3} =\frac {y^3} {(x+3)^3}.$ This implies $x=y$ because the function $f(x):=x^3(x+3)^3$ increases for nonnegative $x$. Taking into account $(x + 1)(y + 1) =4$, we obtain the optimal solution $\{x=1, y=1\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/436066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$p \mid x^2 +n\cdot y^2$ and $\gcd(x,y)=1 \Longleftrightarrow (-n/p) = 1$ Let $n$ be a nonzero integer, let $p$ be an odd prime not dividing $n$. then $ p \mid x^2 + n\cdot y^2$ and $x,y$ co-prime $ \Longleftrightarrow(-n/p) = 1 $
How can i prove this? by $(-n/p)$ i mean the Legendre symbol.
For $\implies$ i have already tried this:
$ p \mid x^2 +n\cdot y^2$, so $x^2 + n\cdot y^2 = 0$ mod $p$. then $x^2 = -n\cdot y^2\mod p$...
So with a little help from my friends this part is done.
Now how to show the other implication?
Greets
Egon
|
Hint: We have $x^2\equiv -ny^2\pmod{p}$. Multiply both sides by $z^2$, where $z$ is the multiplicative inverse of $y$.
Detail: We need to be careful about the statement of the theorem. So we break up the statement and proof into two parts. When we do, we will discover that the result is stated somewhat too informally.
(i) Suppose that $(-n/p)=1$. Then there exist relatively prime integers $x$ and $y$ such that $p$ divides $x^2+ny^2$.
Proof of (i): Since $(-n/p)=1$, by part of the definition of quadratic residue, $n$ is not divisible by $p$. Also, there exists an integer $x$ such that $x^2\equiv -n\pmod{n}$. Thus $x^2+n$ is divisible by $p$, and therefore $x^2+ny^2$ is divisible by $p$, with $y=1$. Note that $x$ and $y$ are relatively prime.
(ii) Suppose there exist relatively prime integers $x$ and $y$ such that $x^2+ny^2$ is divisible by $p$ and $\gcd(x,y)=1$. This is not enough to show that $(-n/p)=1$. For example, let $n=3$. $x=3$, and $y=1$. Thus we must assume in addition that $n$ is not divisible by $p$. We prove the desired result, with the modification that we add in the condition that $n$ is not divisible by $p$.
Proof of (ii): Note that $y$ cannot be divisible by $p$. For if it is, then from $p$ divides $x^2+ny^2$ we conclude that $p$ divides $x^2$. Then $p$ divides $x$, contradicting the fact that $x$ and $y$ are relatively prime.
Since $y$ is not divisible by $p$, it has a multiplicative inverse modulo $p$. That is, there is a $z$ such that $zy\equiv 1\pmod{p}$. Then $x^2z^2+ny^2z^2\equiv 0\pmod{p}$. Thus $(xz)^2\equiv -n\pmod{p}$, and the result follows.
Remark: The theorem should really be stated like this. Let $p$ be an odd prime, and suppose that (the integer) $n$ is not divisible by $p$. Then $(-n/p)=1$ if and only if there exist relatively prime integers $x$ and $y$ such that $x^2+ny^2$ is divisible by $p$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/437720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Proving by induction: $2^n > n^3 $ for any natural number $n > 9$ I need to prove that $$ 2^n > n^3\quad \forall n\in \mathbb N, \;n>9.$$
Now that is actually very easy if we prove it for real numbers using calculus. But I need a proof that uses mathematical induction.
I tried the problem for a long time, but got stuck at one step - I have to prove that:
$$ k^3 > 3k^2 + 3k + 1 $$
Hints???
|
For another way just using $n>9$,
note that when $n=10$, $2^n = 1024 > 1000 = n^3$. Now suppose that $2^n>n^3$ for $n>9$. Then,
$\begin{align*}
2^{n+1} &= 2\cdot2^n \\
&>2n^3 \\
&= n^3 +n^3 \\
&> n^3 + 9n^2 \\
&= n^3 + 3n^2 + 6n^2 \\
&>n^3 + 3n^2 +54n \\
&=n^3+3n^2+3n +51n\\
&>n^3+3n^2+3n+1 \\
&= (n+1)^3.
\end{align*}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/438260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
}
|
How can I find a number $n$ such that $(c + d \times n)$ is a multiple of $(a + n)$? As an example, let $a=5$, $c=246$, and $d=316$. My goal is to find a positive integer $n$ such that
$$(a + n) | (c + d \times n)$$
$$(5 + n) | (246 + 316 \times n)$$
I know that one solution is $18$, since
$$(5 + 18) | (246 + 316 \times 18)$$
$$23 | 5934$$
$$23 \times 258 = 5934$$
The number $24$ is also a solution, as well as a few other numbers.
How can I find solutions to this problem directly? So far, the only way that I have found solutions is through brute force (try every possible number in order), but that is horribly inefficient.
Also, if I were to change the restriction from "being a multiple" to "having a GCF $> 1$" would it make the problem any easier?
|
Make the ansatz
$$d\cdot n + c = (d+k)\cdot(n+a).$$
Expanding the right hand side, you get
$$\begin{align}
d\cdot n + c &= d\cdot n + k\cdot n + d\cdot a + k\cdot a\\
c - d\cdot a &= k\cdot (n+a).
\end{align}$$
So you need to factorise $\lvert c - d\cdot a\rvert$ (if that is $\neq 0$, if $c = d\cdot a$, then all $n$ work), and can then check which factorisations yield a suitable $n$ (the second factor must be $> a$).
In your example, you have $d\cdot a - c = 5\cdot 316 - 246 = 1334 = 2\cdot 23\cdot 29$.
That yields the factorisations
$$\begin{align}
c - d\cdot a &= -1\cdot 1334 \leadsto n = 1334-5 = 1329\\
&= -2\cdot 667 \leadsto n = 667-5 = 662\\
&= -23\cdot 58 \leadsto n = 58-5 = 53\\
&= -29\cdot 46 \leadsto n = 46-5 = 41\\
&= -46\cdot 29 \leadsto n = 29-5 = 24\\
&= -58\cdot 23 \leadsto n = 23-5 = 18\\
&= -667 \cdot 2 \leadsto n = 2-5 = -3 < 0\quad\text{ invalid}\\
&= -1334 \cdot 1 \leadsto n = 1-5 = -4 < 0\quad\text{ invalid}.
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/438330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find all real numbers such that $\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ Find all real numbers such that
$$\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$$
My attempt to the solution :
I tried to square both sides and tried to remove the root but the equation became of 6th degree.Is there an easier method to solve this?
|
Let's multiply
$$
\sqrt{x-1/x} + \sqrt{1 - 1/x} = x\tag{1}
$$
by $(\sqrt{x-1/x} - \sqrt{1 - 1/x})$. Then we get
$$
x-1=x(\sqrt{x-1/x} - \sqrt{1 - 1/x})
$$
$$
\sqrt{x-1/x} - \sqrt{1 - 1/x}=1-1/x\tag{2}
$$
Sum up $(1)$ and $(2)$ to get
$$
2\sqrt{x-1/x}=x-1/x+1
$$
Now make the substitution $y=\sqrt{x-1/x}$. The rest is clear.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/438452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
}
|
Solve the equation $z^3=z+\overline{z}$ I have been trying to solve an equation $z^3=z+\overline{z}$, where $\overline{z}=a-bi$ if $z=a+bi$. But I cant find any clues on how to move forward on that one. Please help.
|
Note that if $z=a+ib$ then $z+\overline{z}=2a$ and
$$
z^3=2a
$$
The cube roots of $2a$ are
\begin{align}
\sqrt[3]{2a}\cdot\omega^{0}
=
&
\sqrt[3]{2a}
\\
\sqrt[3]{2a}\cdot\omega^{1}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)
\\
\sqrt[3]{2a}\cdot\omega^{2}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{2}=\sqrt[3]{2a}\big(\frac{1}{2}-i\frac{\sqrt{3}}{2}\big)
\end{align}
where $\omega=\frac{1}{2}+i\frac{\sqrt[3]{3}}{2}$ is any cubic root of the unit.The possible values of $ a $ and $ b $ are obtained equaling $ z = a + ib$ to the roots of given above $2a$.
\begin{align}
\sqrt[3]{2a}\cdot\omega^{0}
=
&
\sqrt[3]{2a}
\\
\sqrt[3]{2a}\cdot\omega^{1}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{1}
\\
\sqrt[3]{2a}\cdot\omega^{2}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{2}=\sqrt[3]{2a}\big(\frac{1}{2}-i\frac{\sqrt{3}}{2}\big)
\end{align}
More explicitly
\begin{align}
a+ib
=
&
\sqrt[3]{2a}
\\
a+ib
=
&
\frac{1}{2}\sqrt[3]{2a}+i\frac{\sqrt{3}}{2}\sqrt[3]{2a}
\\
a+ib
=
&
\frac{1}{2}\sqrt[3]{2a}-i\frac{\sqrt{3}}{2}\sqrt[3]{2a}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/440000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 1
}
|
How prove this $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge |(a-1)(b-1)(c-1)|+a+b+c$ let $a,b,c$ are positive numbers,and such $abc\le 1$,prove that
$$\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\ge |(a-1)(b-1)(c-1)|+a+b+c$$
|
case 1:
For case $a<1,b<1,c<1$, it is trivial the inequality work(Sudeep's comments)
case 2:
now we prove when one of $a,b,c$ equals $1$,the inequality also works.
WLOG,Let $c=1$, the inequality $\iff b^2+a \ge ab^2+ab$, let $ab=k \le 1 \iff b^3-kb^2-kb+k \ge0 \iff b^3(1-k)+k(b-1)^2(b+1)\ge0$
so the "=" will hold when $k=1,b=1 \implies a=b=c=1$
we will prove the case $abc=1$ first:
there are two cases:
case 3: two of them $\ge 1$ , the inequality $\iff \dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\ge ab+bc+ac \iff a^3b^3-a^2b^3+b^3-b^2-ab+1\ge 0 \iff $
$b^3(a-1)^2(a+1)+(b^2-1)(a-1) \ge 0$ when $a,b$ both $\ge 1 $
$b(ab-1)^2(ab+1)+b^2(1-b)(a^2-1) \ge 0 $ when one of $a,b \ge1$ and another $\le1$
case 4: one of them $\ge1$, the other two $\le1$ ,the inequality $\iff \dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\ge 2(a+b+c)-(ab+bc+ac) \iff a^3b^3+a^2b^3+b^3+b^2+1-2ab^3-2a^2b^2-2b \ge0 \iff (b-1)^2+ab(ab-1)^2+b^3(a-1)^2 \ge 0$
now we discuss $a'b'c'<1$ cases:
let $a'b'c'=k^3,a'=ka,b'=kb,c'=kc,abc=1$, we only consider RHS becasue LHS will be same.
denote $H$ for $a,b,c ,H'$ for $a',b',c'$,
if $H'$ in case 3,then $H$ will be in case 3 also:
$H-H'=(ab+bc+ac)(1-k^2)>0$
if both $H'$ and $H $ in case 4:
$H-H'=2(1-k)(a+b+c)-(1-k^2)(ab+bc+ac)+1-k^3$, let $c \ge 1 \iff $
$2(1-k)(a^2b+b^2a+1)-(1-k^2)(a^2b^2+a+b)+(1-k^3)ab =(1-k)((2ab-(1+k))(a+b)+2)+(1-k)(1+k)(ab-a^2b^2)+(1-k)k^2ab \ge 0 \implies H \ge H'$
if $H'$ in case 4, but $H$ in case 3, then there is always one case that $a=ka'>1$ suppose $c \ge 1, a\ge b$ we can chose $k_2=\dfrac{1}{a'}$,then $a=1,b\le1, c\ge 1$ which make $H$ still in case 4. and $H$ now is belong to case 2.So the inequality is also true.
QED.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/441336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
}
|
Finding $a + b + c$ given that $\;a + \frac{1}{b+\large\frac 1c} = \frac{37}{16}$ Please help me to find the needed sum:
If $a,b,c$ are positive integers such that $\;a + \dfrac{1}{b+\large \frac 1c} = \dfrac{37}{16},\;$
find the value of $\;(a+b+c)$.
Thanks!
|
$$\;\color{blue}{\bf a} + \dfrac{1}{\color{red}{\bf b}+1/\color{green}{\bf c}} = \dfrac{37}{16} = 2 + \dfrac 5{16} = 2 + \dfrac 1{\frac{16}{5}} = \color{blue}{\bf 2} + \dfrac 1{\color{red}{\bf 3} + 1/\color{green}{\bf 5}}$$
Now, match up: $\;\color{blue}{\bf a = \;?}\;\quad \color{red}{\bf b = \;?},\quad \color{green}{\bf c = \;?}$
And then find the sum: $$\quad a + b + c = \;\;?$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/441403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
If $x,y,z>0$ are distinct and $x+y+z=1$ what is the minimum of $\left((1+x)(1+y)(1+z)\right)/\left((1-x)(1-y)(1-z)\right)$? If $x,y,z>0$ are not equal and positive and if $x+y+z=1$ the expression
$$\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}$$
is greater than what quantity?
|
we prove
$$\dfrac{1+x}{1-x}\cdot\dfrac{1+y}{1-y}\cdot\dfrac{1+z}{1-z}\ge 8,x+y+z=1$$
$$\Longleftrightarrow \sum \ln{\left(\dfrac{1+x}{1-x}\right)}\ge 3\ln{2}$$
and we have
$$\ln{\left(\dfrac{1+x}{1-x}\right)}\ge\dfrac{9}{4}x-\dfrac{3}{4}+\ln{2}$$
poof:let
$$G(x)=\ln{\left(\dfrac{1+x}{1-x}\right)}-\left(\dfrac{9}{4}x-\dfrac{3}{4}+\ln{2}\right)$$
$$\Longrightarrow G'(x)=\dfrac{(3x+1)(3x-1)}{4(1-x^2)}$$
since $0<x<1$, then we have
$$G(x)\ge G(\dfrac{1}{3})=0$$
so
$$\sum \ln{\left(\dfrac{1+x}{1-x}\right)}\ge\dfrac{9}{4}(x+y+z)-\dfrac{3}{4}\cdot 3-3\cdot\ln{2}=3\ln{2}$$
by done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/442136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
If $a,b,c > 0$ satisfy $a^2+b^2+c^2=3$ then $\frac{a}{b+c+3}+\frac{b}{a+c+3}+\frac{c}{a+b+3} \geq \frac{3}{5}$
Given $a,b,c > 0$ satisfying the condition $$a^2+b^2+c^2=3,$$
prove that
$$\frac{a}{b+c+3}+\frac{b}{a+c+3}+\frac{c}{a+b+3} \geq \frac{3}{5}.$$
Thank you all
|
The inequality is not TRUE. For example. $a=\sqrt{2}$, b=1, c=0, then
$\frac{\sqrt{2}}{4} + \frac{1}{3+\sqrt{2}} + 0 = 0.3536 + 0.2265 < 0.6$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/443214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Die Roll Probability If Zachary rolls a fair die five times, what is the probability that the sum of his five rolls is 20?
1st I did:
Patterns of 5 that can give us 20
66611, 66521, 65531, 65522, 64442, 64433, 64415,
Where do we go from here?
|
Assuming that you know generating functions.
We are interested in the coefficient of $x^{20}$ in the expansion $(x+x^2+x^3+x^4+x^5+x^6)^5$.
This is equivalent to the coefficient of $x^{15}$ in the expansion $\left( \frac{1- x^6}{1-x} \right)^5 $
The numerator is easily evaluated as
$$ 1- - 5 x^{6} + 10 x^{12} - 10x^{18} + 5 x^{24} - x^{30} $$
The denominator is
$$ \sum x^i {i+4 \choose 4} $$
Hence, the answer is
$$ 10 \times {7 \choose 4} - 5 \times {13 \choose 4} + 1 \times {19 \choose 4} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/447723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
What rational numbers have rational square roots? All rational numbers have the fraction form $$\frac a b,$$ where a and b are integers($b\neq0$).
My question is: for what $a$ and $b$ does the fraction have rational square root? The simple answer would be when both are perfect squares, but if two perfect squares are multiplied by a common integer $n$, the result may not be two perfect squares. Like:$$\frac49 \to \frac 8 {18}$$
And intuitively, without factoring, $a=8$ and $b=18$ must qualify by some standard to have a rational square root.
Once this is solved, can this be extended to any degree of roots? Like for what $a$ and $b$ does the fraction have rational $n$th root?
|
First,
$$
\sqrt{\frac ab}\in\mathbb{Q}\implies\sqrt{ab}=b\,\sqrt{\frac ab}\in\mathbb{Q}\tag{1}
$$
According to this answer, since $\left(\sqrt{ab}\right)^2=ab\in\mathbb{Z}$, if $\sqrt{ab}\in\mathbb{Q}$, then $\sqrt{ab}\in\mathbb{Z}$. Thus, there is a $c\in\mathbb{Z}$ so that
$$
ab=c^2\tag{2}
$$
If $\gcd(a,b)=1$, then, by the Fundamental Theorem of Arithmetic, $(2)$ implies that both $a$ and $b$ are perfect squares.
We can avoid the Fundamental Theorem of Arithmetic by showing that if $\gcd(a,b)=1$, then Bezout's Identity says that because
$$
\begin{align}
1
&=(ax+by)^3\\
&=a^2\left(ax^3+3bx^2y\right)+b^2\left(3axy^2+by^3\right)\tag{3}
\end{align}
$$
we have $\gcd\left(a^2,b^2\right)=1$. Substituting $a\mapsto\frac{a}{\gcd(a,b)}$ and $b\mapsto\frac{b}{\gcd(a,b)}$ into $(3)$ shows that
$$
\gcd\left(a^2,b^2\right)=\gcd(a,b)^2\tag{4}
$$
Thus,
$$
a=a\gcd(a,b)=\gcd\left(a^2,ab\right)=\gcd\left(a^2,c^2\right)=\gcd(a,c)^2\tag{5}
$$
and
$$
b=b\gcd(a,b)=\gcd\left(ab,b^2\right)=\gcd\left(c^2,b^2\right)=\gcd(c,b)^2\tag{6}
$$
Therefore, $a$ and $b$ are perfect squares.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/448172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 8,
"answer_id": 5
}
|
Show that $\frac {\sin(3x)}{ \sin x} + \frac {\cos(3x)}{ \cos x} = 4\cos(2x)$ Show that
$$\frac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x} = 4\cos(2x).$$
|
$$\dfrac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x}$$
$$\dfrac{\sin(3x)\cos x + \cos(3x)\sin x}{\cos x{\sin x}}$$
$$\dfrac {\sin (3x+x)}{\sin x\cos x}$$
$$\dfrac {\sin 4x}{\sin x\cos x}$$
$$\dfrac {2\sin 2x\cos 2x}{\sin x\cos x}$$
$$\dfrac {4\sin x\cos x\cos 2x}{\sin x\cos x}$$
$$4\cos 2x$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/448673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
}
|
How to solve $ (x + 2y - 4)dx - (2x + y - 5)dy = 0$ How to solve the differential equation $(x + 2y - 4)dx - (2x + y - 5)dy = 0$. It's not separable, nor exact nor homogeneous. The solution is $(x - y -1)^3 = C(x + y - 3)$. How can I achieve it?
The other equations similar to this are:
$(1+ x + y)dy - (1- 3x - 3y)dx = 0$
Answer: $3x + y + 2ln(-x - y +1) = k$
$(3x - y + 2)dx + (9x - 3y +1)dy = 0$
Answer: $2x + 6y + C = ln(6x - 2y +1)$
If someone point me out how to solve the first equation I will be likely to solve the others. Thank you very much.
Update: Given Orangutango and Chris help I moved to a solvable d.e. But didn't get the same answer my professor listed. Did I miss some step?
(X + 2Y)dX = (2X+Y)dY
dY/dX = (X + 2Y)/(2X + Y)
Making a substitution to solve the now homogenous:
Y = VX, Y'= V + V'X
V+V'X = (X + 2VX)/(2X + VX)
V+V'X = (1 + 2V)/(2 + V)
V'X = ((1 + 2V)/(2 + V)) - V
V'X = (1 - V^2)/(2 + V)
(2 + V)dV/(1-V^2) = dX/X
Integrating the left side I got:
int (2 + V)dV/1-V^2 =
2 * int dV / (1-V^2) + int V dV / (1-V^2) =
log | (v + 1) / (v-1) | - 1/2 log | V^2 + 1 | + c1
Integrating the right side:
log X + c2
Then replacing V=Y/X and then X=x-2 and Y=y-1 don't seems to yield the proposed answers. Where did my professor got this? Is the above solution correct? Thanks again.
|
$$ (x + 2y - 4)dx - (2x + y - 5)dy = 0 $$
Let $X = x + a$ and $Y= y+ b$ and let's see if we can choose suitable values for $a$ and $b$.
$$ (X - a + 2Y - 2b - 4)dX - (2X-2a + Y - b - 5)dY =0$$
It's easy to find what the "suitable" values should be, since we just want them to cancel out the constants! So we solve the following system of equations:
$$\begin{align*}
-4 &= a + 2b\\
-5 &= 2a + b
\end{align*}$$
I'll let you verify that the correct values are $a=-2$ and $b=-1$. This leaves us with
$$(X + 2Y)dX - (2X+Y)dY=0$$
Can you take it from here?
EDIT: Now let $Y=VX$, and hence $dY = VdX + XdV$.
$$\begin{align*}
X(1 + 2V)dX - X(2+V)(VdX+XdV) &=0\\
X(1-V^2)dX - X^2(2+V)dV &=0\\
X(1-V^2)dX &= X^2(2+V)dV\\
\dfrac{1}{X}dX &= \dfrac{2+V}{1-V^2}dV\\
\log(X) + C_1 &= \dfrac{1}{2}(-3 \log(1-V)+\log(1+V))\\
C_2X^2 &= \dfrac{1+V}{(1-V)^3}\\
C_2X^2(1-V)^3 &= 1+V\\
C_2X^2(1-\dfrac{Y}{X})^3 &= 1+\dfrac{Y}{X}\\
C_2\dfrac{(X-Y)^3}{X}&= 1+\dfrac{Y}{X}\\
C_2(X-Y)^3 &= X + Y\\
C_2(x-y-1)^3 &= x + y - 3
\end{align*}$$
So your teacher has the correct solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/448839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
How prove this $\frac{1}{\cos{A}}-\frac{\sin{\frac{A}{2}}}{\sin{\frac{B}{2}}\sin{\frac{C}{2}}}=4$ in $\Delta ABC $ not An equilateral triangle, let
$$\begin{vmatrix}
2\sin{A}\sin{C}-\cos{B}&2\sin{A}\sin{B}-\cos{C}\\
4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+\cos{B}&4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+\cos{C}
\end{vmatrix}=0$$
prove that
$$\dfrac{1}{\cos{A}}-\dfrac{\sin{\dfrac{A}{2}}}{\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}}=4$$
my idea:
use this well known
$$4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+1=\cos{A}+\cos{B}+\cos{C}$$
then we have
$$\Longrightarrow \begin{vmatrix}
2\sin{A}\sin{C}-\cos{B}&2\sin{A}\sin{B}-\cos{C}\\
\cos{A}+\cos{C}+2\cos{B}-1&\cos{A}+\cos{B}+2\cos{C}-1
\end{vmatrix}=0$$
then
$$(2\sin{A}\sin{C}-\cos{B})(\cos{A}+\cos{B}+2\cos{C}-1)-(2\sin{A}\sin{B}-\cos{C})(\cos{A}+\cos{C}+2\cos{B}-1)=0$$
folowing I can't work,Thank you everyone can help
|
Hint:
Your identity is equivalent to:
$$\frac{\sin(A/2)\sin(B/2)\sin(C/2)}{\cos A} + \sin^2(A/2) = 4\sin(A/2)\sin(B/2)\sin(C/2)$$
$$\frac{1}{2}\left(1-\cos(A)\right)=\left(\cos(A)+\cos(B)+\cos(C)-1\right)\left(1 - \frac{1}{\cos(A)} \right)$$
Notice that this form is quite close to the fraction you get from the determinant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/449215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $(a-d)^2+(b-c)^2\geq 1.6$ if $a^2+4b^2=4$ and $cd=4$ Let $a, b, c, d$ be real numbers. I need to prove this innocent inequality
$$
(a-d)^2+(b-c)^2\geq 1.6
$$
if $a^2+4b^2=4$ and $cd=4$.
I was told that there exist nice and sweet elementary solutions.
|
it is supplementary to Alex's method which more like high school method:
The idea is same,we use two parallel lines to calculate. let slop is $k=-\tan{ \alpha}, 0<\alpha <\dfrac{\pi}{2}$,
the tangent line to ellipse $x^2+4y^2=4$ is :$y=kx+b_1 \to x^2+4(kx+b_1)^2=4, \Delta=0,\implies b_1^2=4k^2+1$
the tangent line to Hyperbola $xy=4$ is:$y=kx+b \to x(kx+b)=4, \Delta=0,\implies k=-\dfrac{b^2}{4^2}$
$b_1=\sqrt{\dfrac{b^4}{64}+1},\cos{\alpha}=\dfrac{1}{\sqrt{1+\tan^2{\alpha}}}=\dfrac{4^2}{\sqrt{b^4+4^4}}$
the distance between two lines is: $D=(b-b_1)\cos{\alpha}=\left(b-\sqrt{\dfrac{b^4}{64}+1}\right)\dfrac{4^2}{\sqrt{b^4+4^4}}=\dfrac{2(8b-\sqrt{b^4+64})}{\sqrt{b^4+4^4}}=\dfrac{2(64b^2-b^4-64)}{\sqrt{b^4+4^4}(b+\sqrt{b^4+64})}=\dfrac{2(960-(b^2-32^2)^2)}{\sqrt{b^4+4^4}(b+\sqrt{b^4+64})}$
the numerator is Parabola, the peak is at $b^2=32,b=4\sqrt{2}$
the denominator is mono increasing function, and it is lead by $b^4$, so it will cause the peak of $D$ move left strongly. The $D(b)$ has two zero points at about $b=1,8$, so the peak point should be around $3$ or $4$( it is guess).
$D(3)=1.3,D(4)=1.25 \implies D^2=1.69>1.6$ so we have at least one solution to show the square of distance of two points $(a,b),(c,d) >1.6 $
if we try $D(3.5)=1.76$, so if we set min is $1.7$,it is much hard to high school student.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/449755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 2
}
|
A divisibility question concerning positive integers Suppose $n$ is a positive integer such that $3n+1$ and $4n+1$ are both perfect squares , then how do we prove that $7|n$ ?
|
Suppose $x^2=3n+1$ and $y^2=4n+1$. Then,
$$
4x^2-3y^2=1
$$
Using continued fractions, we find the solutions to this equation are $(x_k,y_k)$ where
$$
\begin{align}
(x_0,y_0)&=(1,1)\\
(x_1,y_1)&=(13,15)\\
(x_k,y_k)&=14(x_{k-1},y_{k-1})-(x_{k-2},y_{k-2})
\end{align}
$$
Looking at $(x_k,y_k)\text{ mod }7$, we see
$$
\begin{align}
(x_0,y_0)&=(1,1)\\
(x_1,y_1)&=(-1,1)\\
(x_k,y_k)&=-(x_{k-2},y_{k-2})
\end{align}
$$
Thus, $x_k^2\equiv y_k^2\equiv1\pmod{7}$ and therefore, $n\equiv0\pmod{7}$.
Solving Pell's equation with Continued Fractions
When solving $4x^2-3y^2=1$, we want $\frac yx$ to underestimate $\frac2{\sqrt3}$.
The continued fraction for $\frac2{\sqrt3}$ is $(1;\overline{6,2})$:
$$
\begin{array}{c|c}
\frac{2\sqrt3}{3}&2\sqrt3+3&\frac{2\sqrt3+3}{3}&2\sqrt3+3\\
\hline\\
1&6&2&\dots\\
\end{array}
$$
Thus, the convergents are
$$
\begin{array}{c|c}
&&&1&6&2&6&2\\
\hline y_k&0&1&\color{#C00000}{1}&7&\color{#C00000}{15}&97&\color{#C00000}{209}\\
\hline x_k&1&0&\color{#C00000}{1}&6&\color{#C00000}{13}&84&\color{#C00000}{181}\\
\hline k&&&0&1&2&3&4
\end{array}
$$
The underestimates are in red (even $k$).
Due to the alternating continuants, for $k\gt0$, we have
$$
a_{2k}=2a_{2k-1}+a_{2k-2}\quad\text{ and }\quad a_{2k-1}=6a_{2k-2}+a_{2k-3}
$$
which leads to a recursion for the underestimates:
$$
a_{2k}=14a_{2k-2}-a_{2k-4}
$$
where $a_k=(x_k,y_k)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/450254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
}
|
Gaussian Elimination with Scaled Row Pivoting for numerical methods I am solving a system first with basic Gaussian Elimination, and then Gaussian Elimination with scaled row pivoting (used in numerical methods)
Basic Gaussian Elimination on the system $Ax=b$:
\begin{equation}
\begin{pmatrix}-1& 1& -4 \\
2& 2& 0 \\
3& 3& 2 \end{pmatrix}
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =
\begin{pmatrix}0\\1\\\frac{1}{2}\end{pmatrix}
\end{equation}
Let $A_i$ denote the $i^{th}$ row of matrix $A$ and let $A^{(1)} A^{(2)}...$ denote the matrix after the first, second and so forth elementary row operations. Note that
$A^{0} =A$.
Compute the following elementary row operations:
\begin{align}
A^{(1)}_2 =& A^{(0)}_2 - (-2)A^{(0)}_1 \\
A^{(1)}_3 =& A^{(0)}_3 - (-3)A^{(0)}_1
\end{align}
This yields:
This yields:
\begin{equation}
\begin{pmatrix}-1& 1& -4 \\
0& 4& -8 \\
0& 6& -10 \end{pmatrix}
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =
\begin{pmatrix}0\\1\\-1\end{pmatrix}
\end{equation}
Compute:
\begin{equation} A^{(2)}_3 = A^{(1)}_3 - (\frac{3}{2})A^{(1)}_2\end{equation}
This yields:
\begin{equation}
\begin{pmatrix}-1& 1& -4\\
0& 4& -8\\
0& 0& 2\end{pmatrix}
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=
\begin{pmatrix}0\\1\\-1\end{pmatrix}
\end{equation}
Thus we have:
\begin{equation} x=\begin{pmatrix}\frac{5}{4}\\
\frac{-3}{4}\\
\frac{-1}{2}\end{pmatrix}\end{equation}
Now I will solve the same system with Scaled Row Pivoting. The $i^{th}$ element of the list $S$ will denote the maximum element in row $i$ in matrix $A$. $P$ will denote the order of the rows. Initially we have:
\begin{equation}
S = (4, 2, 3) \\
P = (2, 1, 3)
\end{equation}
Swap rows $1$ and $2$ since row $2$ has the maximum pivot relative to its row:
\begin{equation}
\begin{pmatrix}2&2&0\\
-1&1&-4\\
3& 3& 2\end{pmatrix}
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =
\begin{pmatrix}1\\0\\\frac{1}{2}\end{pmatrix}
\end{equation}
Now compute the following elementary row operations w.r.t the ordering given by $p$:
\begin{align}
A^{(1)}_1 =& A^{(0)}_1 - (\frac{-1}{2})A^{(0)}_2 \\
A^{(1)}_3 =& A^{(0)}_3 - (\frac{3}{2})A^{(0)}_2
\end{align}
This yields:
\begin{equation}
\begin{pmatrix}2&2&0\\
0&2&-4\\
0&0&2\end{pmatrix}
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=
\begin{pmatrix}1\\\frac{1}{2}\\-1
\end{pmatrix}
\end{equation}
Now using back substitution to solve for $x$ we get:
\begin{equation}
x=\begin{pmatrix}\frac{-1}{4}\\\frac{3}{4}\\\frac{-1}{2}\end{pmatrix}
\end{equation}
Clearly, I must have made a mistake along the way since the solutions for both methods are not the same! I know that the scaled pivoting is incorrect as I checked my solution in a CAS and it matched the solution for the Basic Method.
Please show me what I have done wrong in the scaled pivoting algorithm.
|
You miscomputed $x_2$ in the back substitution of the row-pivoted system, that's the origin of the discrepancy.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/454356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Solve for c , $y = x + c \big( \frac{mx}{c} + s \big)^a$ I get this equation
$y = x + c \big( \frac{mx}{c} + s \big)^a$
how can I get the $c$ or $m$ ?
I try with $\ln$
$\ln\big(\frac{y-x}{c}\big) = a \ln \big( \frac{mx}{c} + s\big)$
and now ?
|
Here is how you solve for $m$:
\begin{align*}
\frac{y-x}{c}=\left(\frac{mx}{c}+s\right)^{a}\\
\left(\frac{y-x}{c}\right)^{\frac{1}{a}}=\left(\frac{mx}{c}+s\right)\\
\left(\frac{y-x}{c}\right)^{\frac{1}{a}}-s=\frac{mx}{c}\\
\end{align*}
Hence
\begin{align*}
m=\frac{c}{x}\left(\left(\frac{y-x}{c}\right)^{\frac{1}{a}}-s\right)
\end{align*}
I think to solve $c$, you have to use numerical method since it is not explicit.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/456919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
The values of $N$ for which $N(N-101)$ is a perfect square
For how many values of $N$ (integer), $N(N-101)$ is a perfect square number?
I started in this way.
Let $N(N-101)=a^2$
or $N^2-101N-a^2=0.$
Now if the discriminant of this equation is a perfect square then we can solve $N$.
But I can't progress further.
|
We have $$N^2-101N = a^2 \implies (N-101/2)^2 - (101/2)^2 = a^2 \implies (N-101/2)^2 - a^2 = (101/2)^2$$
This gives us
$$(2N-101)^2 - (2a)^2 = 101^2 \implies (2N+2a-101)(2N-2a-101)=101^2$$
Note that $101^2$ can be written as
$$101 \times 101 = 1 \times 101^2 = 101^2 \times 1 = (-101) \times (-101) = (-1) \times (-101^2) = (-101^2) \times (-1)$$
This gives us
$$a=0 \text{ and }N = 0,101$$
$$4N - 202 = 1+101^2 \implies 4N = 10202+202 \implies N = \dfrac{10404}4 = 2601$$
$$4N - 202 = -1-101^2 \implies 4N = -10202+202 \implies N = -2500$$
Hence, the possible values of $N$ are
$$N=0,101,-2500,2601$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/460022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
}
|
Initial value problem differential equation $y' = (x-1)(y-2)$ $$y' = (x-1)(y-2)$$
$y(2)= 4$
$$\frac{1}{y-2}dy = (x-1)dx$$
$$\int \frac{1}{y-2}dy =\int (x-1)dx$$
$$\ln(y-2) = \frac{x^2}{2} - x + c$$
$$y - 2 = e^{\frac{x^2}{2} - x + c} $$
$$y = e^{\frac{x^2}{2} - x + c} + 2$$
plug in the inital value
$$y = e^{\frac{2^2}{2} - 2 + c} + 2$$
$$y = e^c+ 2$$
I feel like this is wrong anyways, so where did I go wrong?
|
It would be a good remark that in the step you wrote:
$$\ln(y-2)=x^2/2-x+c$$ we inserted this strong point: $$\ln|y-2|=x^2/2-x+c, ~~y\neq 2$$
Now, if we assume to find just one solution and having $y>2$, then $$y=\exp(x^2/2-x+c)+2=C\exp(x^2/2-x)+2$$ and therefore when $x=2$ then $y=4$ makes the latter equality:
$$4=C\exp(2-2)+2\longrightarrow C=2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/460608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Scaling points in circle I have a set of points that fall inside of a circle with a radius of $1$ and a center of $(0, 0)$. I want to know how to scale those points so that they all have a radius between $0.5$ and $0.75$.
For example, if I have $x = 1$, $y = 0$ then $x_1 = 0.75$, $y_1 = 0$. But that is an easy case on the circumference. What would $x = 0.1$, $y = -0.1$ be and how would you find it?
|
For any point $(x,y)$, consider the transformation:
$$
f(x,y)=\left(0.25+\dfrac{0.5}{\sqrt{x^2+y^2}}\right)(x,y)
$$
Notice that the distance from this new point to the origin is:
$$
\left(0.25+\dfrac{0.5}{\sqrt{x^2+y^2}}\right)\sqrt{x^2+y^2}=0.25\sqrt{x^2+y^2}+0.5
$$
Hence, since $0 \leq \sqrt{x^2+y^2} \leq 1$, it follows that $0 \leq 0.25\sqrt{x^2+y^2} \leq 0.25$ which implies that $0.5 \leq 0.25\sqrt{x^2+y^2}+0.5 \leq 0.75$. Indeed, observe that this mapping yields:
$$(1,0) \to \left(0.25+\dfrac{0.5}{\sqrt{1^2+0^2}}\right)(1.0)=0.75(1,0)=(0.75,0) \\$$
as well as
$$ \begin{align*}
(0.1, -0.1) \to \left(0.25+\dfrac{0.5}{\sqrt{0.1^2+(-0.1)^2}}\right)(0.1, -0.1) &=\left(\dfrac14+\dfrac5{\sqrt2} \right)(0.1, -0.1) \\
&\approx(0.37855,-0.37855) \\
\end{align*} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/462254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find $a$ such that $P(X\le a)=\frac 1 2$
Given probability density function $f(x)=\begin{cases} 1.5(1-x^2),&0<x<1\\0& \mbox{otherwise}\end{cases}$, calculate for which '$a$', $P(X\le a)=\frac 1 2$ (the solution is $2\cos\left(\frac{4\pi}{9}\right))$.
I calculated the integral and got $P(X\le a)=\displaystyle\int_0^af_X(t)dt=1.5a\left(1-\frac {a^2}{3}\right) \stackrel?= \frac 1 2$. I substituted $a=\cos\theta$ and then got $$\frac 1 2=1.5\cos\theta-0.5\cos^3\theta$$$$1=3\cos\theta -\cos\theta(\cos^2\theta)=3\cos\theta-\cos\theta(\cos^2\theta)=3\cos\theta-\cos\theta\left(\frac {1-\cos 2\theta} 2\right)$$ and then using some identites I got $$-2.75\cos\theta-0.25\cos3\theta+1=0$$ or after multiplying by (-4 )$$11\cos\theta+\cos3\theta-4=0$$ How can I solve this equation?
|
We have the equation $a^3-3a+1=0$. Of course it can be solved numerically. But we show how cosines get into the game.
Let $a=2b$. The equation becomes $8b^3-6b+1=0$, or equivalently $4b^3 -3b=-\frac{1}{2}$.
Recall the identity $\cos 3\theta=4\cos^3\theta-3\cos\theta$. So we can rewrite our equation as $\cos3\theta=-\frac{1}{2}$. This has $3$ solutions, the primary one being $3\theta=\frac{2\pi}{3}$. But we also have the possibilities $3\theta=\frac{4\pi}{3}$ and $\frac{8\pi}{3}$. That gives $\theta$ equal to one of $\frac{2\pi}{9}$, $\frac{4\pi}{9}$ and $\frac{8\pi}{9}$.
Now we need to pick out the one that works. This turns out to be $\theta=\frac{4\pi}{9}$, giving solution $a=2\cos\left(\frac{4\pi}{9}\right)$.
Remark: Amusing! It is not the first time I have done a calculation much like this one, except for using $-\frac{1}{2}$ instead of $\frac{1}{2}$. With that small change, it is part of the standard argument that one cannot trisect the $60^\circ$ angle with straightedge and compass. Had never imagined giving the calculation as part of the solution of a "probability" problem!
The "trigonometric method" for solving cubics with $3$ real roots (the casus irreducibilis) is due to François Viète.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/463368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
interesting Integral , alternative solution. Show the following relation:
$$\int_{0}^{\infty} \frac{x^{29}}{(5x^2+49)^{17}} \,\mathrm dx = \frac{14!}{2\cdot 49^2 \cdot 5^{15 }\cdot 16!}.$$
I came across this intgeral on a physics forum and solved by (1) making a substitution (2) finding a recursive formula (with integration by parts).
I would like to see a clever alternative solution to this one, since the only way I would solve such an integral is by finding a recursive formula. But I'm sure there are other ways/methods to do it (maybe more elegant) that I want to discover.
|
$$I_{(n,m)}=\int_0^\infty \frac{x^n}{(ax^2+b)^m} \,\mathrm dx$$
$$=x^{n-1}\int_0^\infty\frac{xdx}{(ax^2+b)^m}-\int_0^\infty\left(\frac{d(x^{n-1})}{dx}\int\frac{xdx}{(ax^2+b)^m}\right)dx $$
$$=x^{n-1}\frac1{-2a(m-1)(ax^2+b)^{m-1}}\big|_0^\infty-\frac{n-1}{-2a(m-1)}\int_0^\infty \frac{x^{n-2}}{(ax^2+b)^{m-1}}dx$$
$\lim_{x\to\infty}x^{n-1}\frac1{(ax^2+b)^{m-1}}=\lim_{x\to\infty}\frac1{aO(x^{2m-2-(n-1)})}=0$ if $2m-2-(n-1)=2m-n-1\ge1\iff n\le 2m-2$
$$\implies I_{(n,m)}=\frac{n-1}{2a(m-1)}I_{(n-2,m-1)}\text{if } n\le 2m-2$$
We have $n=29,m=17$ So, we start with $n=2m-5<2m-2$
$$\implies I_{(2m-5,m)}=\frac{2m-5-1}{2a(m-1)}I_{(2m-5-2,m-1)}=\frac{2m-3}{a(m-1)}I_{(2m-7,m-1)}$$
Replacing $m$ with $m-1,$
$$\implies I_{(2m-5,m)}=-\frac{2m-5-1}{2a(m-2)}I_{(2m-5-2,m-1)}=-\frac{m-3}{a(m-2)}I_{(2m-7,m-1)}$$
$$m=17\implies I_{(29,17)}=-\frac{14}{16a}I_{(27,16)}$$
$$m=16\implies I_{(27,16)}=-\frac{13}{15a}I_{(25,15)}$$
$$\cdots$$
$$m=4\implies I_{(3,4)}=-\frac1{3a}I_{(1,3)}$$
Now, $$I_{(1,3)}=\int_0^\infty \frac x{(ax^2+b)^3}dx=\frac1{2a}\int_b^\infty \frac {du}{u^3}=\frac{1}{2a(-2)u^2}\big|_b^\infty=\frac1{4ab^2}$$
$$\implies I_{(29,17)}=\frac{14}{(16\cdot15\cdots4\cdot3)\cdot a^{14}\cdot 4ab^2}=\frac{14!2!}{16!\cdot4a^{15}b^2}=\frac{14!}{16!\cdot2a^{15}b^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/464181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
}
|
Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7) $ The question is to simplify $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$ without using a calculator .
My friend has given me this challenge . I solved it by expanding $$(a+b+c)(-a+b+c)(a-b+c)(a+b-c) = -a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4$$ and then substituting $a,b,c=\sqrt 5 , \sqrt6 , \sqrt7$ respectively to get the answer $104$ .
But I suppose there is a more elegant and easy way to solve this problem .
Can anyone find it ?
|
As $$(a+b+c)(-a+b+c)=\{(b+c)+a\}\{(b+c)-a\}=(b+c)^2-a^2,$$
$$(\sqrt5 +\sqrt6 +\sqrt7)(−\sqrt5+\sqrt6+\sqrt7)$$
$$=(\sqrt6+\sqrt7)^2-(\sqrt5)^2=6+7+2\sqrt7\cdot\sqrt6-5=8+2\sqrt{42}=2\sqrt{42}+8$$
Again as $$(a-b+c)(a+b-c)=\{a+(b-c)\}\{a-(b-c)\}=a^2-(b-c)^2,$$
$$(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)=\{\sqrt5 − (\sqrt6-\sqrt7)\}\{\sqrt5 + (\sqrt6-\sqrt7)\}$$
$$=(\sqrt5)^2-(\sqrt6 − \sqrt7)^2=5-(6+7-\sqrt7\cdot\sqrt6)=2\sqrt{42}-8$$
So, the product $$=(2\sqrt{42}-8)(2\sqrt{42}+8)=(2\sqrt{42})^2-8^2=4\cdot42-64=104$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/465103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 4
}
|
Factorise: $2a^4 + a^2b^2 + ab^3 + b^4$ Factorize : $$2a^4 + a^2b^2 + ab^3 + b^4$$
Here is what I did:
$$a^4+b^4+2a^2b^2+a^4-a^2b^2+ab^3+b^4$$
$$(a^2+b^2)^2+a^2(a^2-b^2)+b^3(a+b)$$
$$(a^2+b^2)^2+a^2(a+b)(a-b)+b^3(a+b)$$
$$(a^2+b^2)^2+(a+b)((a^2(a-b)) +b^3)$$
$$(a^2+b^2)^2+(a+b)(a^3-a^2b+b^3)$$
At this point I don't know what to do and am feeling that my direction is wrong. Please help me.
( Wolfram alpha says that the answer is $(a^2-a b+b^2) (2 a^2+2 a b+b^2)$ but how? )
|
Just Simple Factorisation:
$2a^4+a^2b^2+ab^3+b^4$
$=2a^4+2ab^3+a^2b^2−ab^3+b^4$
$=2a(a^3+b^3)+a^2b^2−ab^3+b^4$
$=2a(a+b)(a^2-ab+b^2)+b^2(a^2-ab+b^2)$
$=(a^2-ab+b^2)(2a^2+2ab+b^2)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/465136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
}
|
Prove the Inequality: $\sum\frac{x^3}{2x^2+y^2}\ge\frac{x+y+z}{3}$ Let $x, y, z>0$. Prove that:
$$\frac{x^3}{2x^2+y^2}+\frac{y^3}{2y^2+z^2}+\frac{z^3}{2z^2+x^2}\ge\frac{x+y+z}{3}$$
|
See my solution from 2006 here:
https://artofproblemsolving.com/community/c6h22937p427220
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/468925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
If $\sin(a)\sin(b)\sin(c)+\cos(a)\cos(b)=1$ then find the value of $\sin(c)$ If $$\sin(a)\sin(b)\sin(c)+\cos(a)\cos(b)=1,$$;where abc are the angles of the triangle.!
then find the value of $\sin(c)$. By trial and error put this triangle as right angled isosceles and got the answer..!! But i want a complete proof!
|
Consider the two unit vectors on $S^2$ with spherical polar coordinates
$(a, 0)$ and $(b,\frac{\pi}{2}-c)$, its components in $\mathbb{R}^3$ are:
$$\begin{align}&(\sin a, 0, \cos a)\\
\text{ and }\quad&(\sin b \cos(\frac{\pi}{2}-c), \sin b\sin(\frac{\pi}{2}-c), \cos b)
=(\sin b\sin c,\sin b\cos c,\cos b)
\end{align}$$
The angle $\theta$ between them satisfies:
$$\cos\theta = \cos a\cos b + \sin a\sin b\sin c$$
Geometrically, $\cos\theta = 1$ if and only if these two unit vector coincides.
This in turn implies $a = b$ and $c = \frac{\pi}{2}$. By sine law, the two sides opposites to
angle $a$, $b$ have equal length. So the triangle is an right angled isosceles triangle and $\sin c = 1$.
Alternatively, one can use the identity
$$\begin{align} & (\sin a-\sin b \sin c)^2+(\sin b\cos c)^2 + (\cos a-\cos b)^2\\
= & 2 \left( 1 - ( \cos a\cos b + \sin a \sin b\sin c)\right)
\end{align}$$
to conclude $\cos c = 0$ and hence $\sin c = 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/469599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
A $4$ variable inequality If $a,b,c,d$ are positive numbers such that $c^2+d^2=(a^2+b^2)^3$, prove that
$$\frac{a^3}{c} + \frac{b^3}{d} \ge 1,$$
with equality if and only if $ad=bc$.
Source: Don Sokolowsky, Crux Mathematicorum, Vol. 6, No. 8, October 1980, p.259.
|
Cauchy-Schwarz gives
$$\left( \frac{a^3}{c} + \frac{b^3}{d}\right)(ac+bd)\ge (a^2+b^2)^2.$$
So it suffices to show that
$$(a^2+b^2)^2 \ge ac + bd.$$
Squaring, this is equivalent to showing
$$(a^2+b^2)^3(a^2+b^2) \ge(ac+bd).$$
This follows immediately from the hypothesis that $(a^2+b^2)^3=c^2+d^2$ and Cauchy-Schwarz.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/471739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.