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Problem solving limit infinity/infinity. I cannot solve this limit:
$$\lim_{x\rightarrow\infty}
\frac {(3x^2-4) \left( \sqrt [3]{2x^2+1}+1
\right)^2}{ (2x-1) \left( 4-\sqrt {8x^3-2}
\right)x^{3/2}}$$
I make this:
$${\frac { \left( 3\,{x}^{2}-4 \right) \left( \sqrt [3]{2\,{x}^{2}+1}+1
\right) ^{2}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) {x}^{3/2}}} =3\,{\frac {\sqrt {x} \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-
1 \right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) }}-4\,{\frac {
\left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-1 \right) \left( 4-
\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}+6\,{\frac {\sqrt {x}\sqrt [3]
{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) }}-8\,{\frac {\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1
\right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}+3\,{\frac
{\sqrt {x}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) }}-4\,{\frac {1}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{
x}^{3}-2} \right) {x}^{3/2}}}$$
then
$$\lim_{x\rightarrow\infty}3\,{\frac {\sqrt {x} \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-
1 \right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-4\,{\frac { \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-1
\right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}
$$
$$\lim_{x\rightarrow\infty}6\,{\frac {\sqrt {x}\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right)
\left( 4-\sqrt {8\,{x}^{3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-8\,{\frac {\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right) \left( 4-
\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}
$$
$$\lim_{x\rightarrow\infty}3\,{\frac {\sqrt {x}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{
3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-4\,{\frac {1}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) {x}^{3/2}}}
$$
but I cannot solve this
NOTE: I cannot use L'hopital for finding this limit.
| $\lim_\limits{x\rightarrow\infty}
{\frac { \left( 3\,{x}^{2}-4 \right) \left( \sqrt [3]{2\,{x}^{2}+1}+1
\right) ^{2}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) {x}^{3/2}}}$
$\lim_\limits{x\rightarrow\infty}
{\frac { \left( 3\,{x}^{2}-4 \right) \left( (2x^2+1)^{1/3}+1
\right) ^{2}}{ \left( 2\,x-1 \right) \left( 4-(8x^3-2)^{1/2}
\right) {x}^{3/2}}}$
find the highest power of x in the numerator and the denominator.
$\lim_\limits{x\to\infty}
\frac{3*2^{2/3}x^{10/3} + ax^n + bx^m... }{2*8^{3/2}x^4+cx^{p}+dx^q..} $
with $m <n < 10/3$ and $p<q < 4$
Divide top and bottom by $x^4$
$\lim_\limits{x\to\infty}
\frac{3*2^{2/3}x^{-2/3} + ax^{n-4} + bx^{m-4}... }{2*8^{3/2}+cx^{p-4}+dx^{q-4}...} $
as x goes to infinity all the terms with x to a negative power go to 0, leaving a zero in the numerator and a constant in the denominator.
$\lim_\limits{x\to\infty}
\frac{3*2^{2/3}x^{-2/3} + ax^{n-4} + bx^{m-4}... }{2*8^{3/2}+cx^{p-4}+dx^{q-4}...}=0 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For real numbers $a,b,c$ calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$ if we have... For real numbers $a,b,c$ we have: $a+b+c=11$ and $\frac1{a+b}+\frac1{b+c}+\frac1{c+a}=\frac{13}{17}$, calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$
I think we should use a trick to solve this,because doing algebraic operations on this problem are too tedius!
| $\frac{c}{a+b} +\frac{a}{b+c}+\frac{b}{c+a}=\frac{11-a-b}{a+b}+\frac{11-b-c}{b+c}+\frac{11-a-c}{c+a}
=(11\cdot \frac{13}{17})-3=\frac{92}{17}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does $\cos(x)\cdot\cos\left(\frac{3}{2}x\right)$ become $\frac{1}{2}\left(\cos\left(\frac{1}{2}x\right) + \cos\left(\frac{5}{2}x\right)\right)$? How can you rewrite $\cos(x)\cdot\cos\left(\frac{3}{2}x\right)$ to $\frac{1}{2}\left(\cos\left(\frac{1}{2}x\right) + \cos\left(\frac{5}{2}x\right)\right)$?
What rules have been used? I need it on this form to compare it with a fourier series.
| Hint. One may use
$$\begin{align}
\cos a \cos b &=\frac12\left((\cos (a-b)+\cos (a+b)\right).
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the set of values of x for which $\frac{x+1}{2x-3}<\frac{1}{x-3}$ Here's what I've done:
$\frac{x+1}{2x-3}<\frac{1}{x-3}$
$x+1<\frac{2x-3}{x-3}$
$(x+1)(x-3)<2x-3$
$x^2-2x-3<2x-3$
$x^2-4x<0$
$x(x-4)<0$
$0<x<4$
However this clearly fails because when $x$ is $2$, for example, the inequality fails.
Where have I gone wrong?
| Beware of cross-multiplying when solving inequalities! You might be multiplying by a negative number – which would invalidate the inequality.
One way to proceed is to bring everything to one side and simplify:
$\quad\quad\quad\dfrac{x+1}{2x+3}\ <\ \dfrac1{x-3}$
$\implies\ 0\ <\ \dfrac1{x-3}-\dfrac{x+1}{2x+3}$
$\implies\ 0\ <\ \dfrac{2x+3-(x+1)(x-3)}{(x-3)(2x+3)}$
…etc.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Differential equation system IVP appears to be wrong $Y(0) = (2,5)$ Let $A = \Bigg ( \begin{array}{cc} 1 & -1 \\ 1 & 3 \end{array} \Bigg )$
Need to solve the IVP: $(x(0),y(0)) = (2,5)$
Solving for the eigenvalues get the characteristic polynomial:
$\lambda^2 - 4 \lambda +4 \implies \lambda_1=\lambda_2 = 2$
for the eigenvector I get $(1,-1)$
so as a general solution I should have $Y(t) = k_1e^{2t}(1,-1) + k_2e^{2t}(1,-1)$ but trying to solve for the initial condition given I am getting that it's not a possible solution.
Am I missing something?
| I will give you a solution that uses matrix exponential of the jordan normal form. Generally, the solution of $y(t)=Ay(t), y(0)=(2,5)^{\top}=y_0$ is given by $y(t)=e^{At}y_0$, so we must focus on $e^{At}$. You already remarked that the characteristic equation has one root $\lambda=2$, but the corresponding eigenspace only dimension $\dim E_{\lambda=2}=1$ with
\begin{align}
E_{\lambda=2}=\ker(A-2I)=\ker\begin{pmatrix} -1 &-1 \\
1 &1
\end{pmatrix}=\left\langle \begin{pmatrix} 1\\ -1\end{pmatrix}\right\rangle
\end{align}
Let $v=(1,-1)^{\top}$ be the eigenvector of $A$ to $\lambda=2$. From $(A-2I)w=v$ we find a generalized eigenvector $w=(-1,0)^{\top}$. Since $A$ isn't diagonalizable, we know that the jordan normal form is
\begin{align}
J=
\begin{pmatrix}
2 & 1\\
0 & 2
\end{pmatrix}
\end{align}
Now we define
\begin{align}
P=\begin{pmatrix}
1 & -1\\
-1 & 0
\end{pmatrix}\qquad \Rightarrow\qquad
P^{-1}=
\begin{pmatrix}
0 & -1\\
-1 & -1
\end{pmatrix}
\end{align}
From that we get
\begin{align}
PJP^{-1}=\begin{pmatrix}
1 & -1\\
-1 & 0
\end{pmatrix}\begin{pmatrix}
2 & 1\\
0 & 2
\end{pmatrix}\begin{pmatrix}
0 & -1\\
-1 & -1
\end{pmatrix}=\begin{pmatrix}
1 & -1\\
1 & 3
\end{pmatrix}=A
\end{align}
This allows us to write
\begin{align}
\exp(At)=\exp(PJP^{-1}t)=P\exp(Jt)P^{-1}
\end{align}
The next step is to calculate $\exp(Jt)$. Maybe you find in some books or your lecture notes that this is
\begin{align}
\exp(Jt)=e^{2t}\begin{pmatrix}
1 & t\\
0 & 1
\end{pmatrix}=\begin{pmatrix}
e^{2t} & te^{2t}\\
0 & e^{2t}
\end{pmatrix}
\end{align}
You will then have
\begin{align}
y(t)=P\exp(Jt)P^{-1}y_0=
\begin{pmatrix}
(1-t)e^{2t} & -te^{2t}\\
te^{2t} & (1+t)e^{2t}
\end{pmatrix}\begin{pmatrix}
2\\ 5
\end{pmatrix}
=\begin{pmatrix}
2(1-t)e^{2t}-5te^{2t}\\ 2te^{2t}+5(t+1)e^{2t}
\end{pmatrix}
\end{align}
This was the first answer i gave...
The first solution is standard. An eigenvector of $\lambda=2$ is $v=(1,-1)^{\top}$, so $Y_1(t)=k_1e^{2t}v$. Since now the Eigenspace of $\lambda=2$ has dimension $\dim E_{\lambda}=1$, we need another linear independent solution. That is
\begin{align}
Y_2(t)=k_2e^{2t}(vt+w)
\end{align}
with a solution $w$ of $(A-2I)w=v$. (Take $Y_2$ into the differential equation an you will get this). That is $w=(-1,0)^{\top}$. The general solution is now
\begin{align}
Y(t)=k_1e^{2t}\binom{1}{-1}+k_2te^{2t}\binom{1}{-1}+k_2e^{2t}\binom{-1}{0}
\end{align}
The constants are determined by $Y(0)=(2,5)^{\top}$ and i hope you will get the same solution as above.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $ \int \frac{1+x\cos(x)}{x(1-x^2(e^{2\sin(x)}))}dx $ $$ \int \frac{1+x\cos x}{x(1-x^2e^{2\sin x})}dx $$
Attempt:
I substituted $(1-xe^{2\sin(x)})$ by $u$ and tried from there by differentiating it. But I get stuck midway.
| HINT:
Note that we can write
$$\begin{align}
\frac{1+x\cos(x)}{x(1-x^2e^{2\sin(x)})}&=(1+x\cos(x))\left(\frac{1}{x(1-x^2e^{2\sin(x)})}\right)\\\\
&=(1+x\cos(x))\left(\frac{1}{x}+\frac{xe^{2\sin(x)}}{1-x^2e^{2\sin(x)}}\right)\\\\
&=\frac1x+\cos(x)+\frac{(x+x^2\cos(x))e^{2\sin(x)}}{1-x^2e^{2\sin(x)}} \tag 1
\end{align}$$
Can you find a substitution for the third term in $(1)$ that facilitates quick integration?
| {
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Rational ODE $y'=\frac{3x^2-2xy+2}{6y^2-x^2+3}$ I really don't know how to start to solve this ODE:
$$y'=\frac{3x^2-2xy+2}{6y^2-x^2+3}$$
I know that somehow I have to isolate $y$ but how?
| Hint:
Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=164:
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\dfrac{3x^2-2x^2u+2}{6x^2u^2-x^2+3}$
$x\dfrac{du}{dx}=\dfrac{(3-2u)x^2+2}{(6u^2-1)x^2+3}-u$
$x\dfrac{du}{dx}=\dfrac{(3-u-6u^3)x^2-3u+2}{(6u^2-1)x^2+3}$
Let $v=x^2$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=2x\dfrac{du}{dv}$
$\therefore2x^2\dfrac{du}{dv}=\dfrac{(3-u-6u^3)x^2-3u+2}{(6u^2-1)x^2+3}$
$2v\dfrac{du}{dv}=\dfrac{(3-u-6u^3)v-3u+2}{(6u^2-1)v+3}$
$((3-u-6u^3)v-3u+2)\dfrac{dv}{du}=2(6u^2-1)v^2+6v$
This belongs to an Abel equation of the second kind.
| {
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Proving by induction that $n! < (\frac{n+1}{2})^n$ As an analysis homework I have to prove by induction that
$n! < (\frac{n+1}{2})^n : (2 \le n \in\mathbb{N})$
For $n = 2$ this is trivial, but for $n+1$ no matter how I transform the equation I can't seem to get $(\frac{n+2}{2})^{n+1}$ on the right-hand side. I'm sure that this is an easy homework and I'm missing something blatantly obvious but the closest I got was $(n+1)! < \frac{(n+2)^{n+1}}{2^{n}}$.
Any help is greatly appreciated.
| Without induction
and completely elementary:
$n!
= \prod_{k=1}^n k
$
so
$\begin{array}\\
n!^2
&= (\prod_{k=1}^n k)^2\\
&= (\prod_{k=1}^n k)(\prod_{k=1}^n k)\\
&= (\prod_{k=1}^n k)(\prod_{k=1}^n (n+1-k))\\
&= \prod_{k=1}^n k(n+1-k)\\
&= \prod_{k=1}^n (k(n+1)-k^2)\\
&= \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}-\dfrac{(n+1)^2}{4}+k(n+1)-k^2\right)\\
&= \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}-\left(\dfrac{(n+1)^2}{4}-k(n+1)+k^2\right)\right)\\
&= \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}-\left(\dfrac{n+1}{2}-k\right)^2\right)\\
&> \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}\right)\\
&= \left(\dfrac{(n+1)^2}{4}\right)^n\\
&= \left(\dfrac{n+1}{2}\right)^{2n}\\
\text{so}\\
n!
&> \left(\dfrac{n+1}{2}\right)^{n}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
} |
write down the expression for $\sin (15°)$ using the double angle formula. show that $\sin 15^\circ=\frac {\sqrt3 -1}{2\sqrt2}$ using $\cos2A=1-2\sin^2A$
However I got $\sin 15^\circ= \sqrt{\frac {2-\sqrt 3}{4}}$ instead.
| $cos(2A)$ = $1 - 2Sin^2(A)$
$\Rightarrow$ $cos(30)$ = $1 - 2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3}{2}$ = $1 - 2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3}{2}$ - 1 = $-2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3 - 2}{2}$ = $-2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3 - 2}{4}$ = $-Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt{2 - \sqrt3}}{2}$ = $Sin(15)$
$\Rightarrow$ $\sqrt{\frac{4 - 2\sqrt3}{8}}$ = $Sin(15)$
$\Rightarrow$ $\sqrt{\frac{(\sqrt3 - 1)^2}{8}}$ = $Sin(15)$
$\Rightarrow$ $\frac{(\sqrt3 - 1)}{2\sqrt2}$ = $Sin(15)$
$\Rightarrow$ $\frac{(\sqrt3 - 1)}{2\sqrt2}$ = $Sin(15)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$12^7+8^8$ divided by $13$ I Need to find what the remainder is when $12^7+8^8$ is divided by $13$
I have a solution, but don't know if it is right.
$12=-1\mod13$
$12^7=-1\mod13$
$8=8\mod13$
$8^2=6\mod13$
$8^4=10\mod13$
$8^8=9\mod13$
Then I did $-1\mod13+9\mod13=8\mod13$, so the remainder is $8$.
If someone could tell me if this is correct, it would be appreciated. Thanks
| You are quite incorrect in your calculations. $8^2 \equiv -1 \pmod {13}$ in your second line.
However, there is an easier way you can proceed. $$12^{7}+8^8 \equiv (-1)^7+2^{24} \equiv -1+(2^{12})^2 \equiv -1+1 \pmod {13}$$ since $2^{12} \equiv 1 \pmod{13}$ from Fermat's Little Theorem.
| {
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"source": "stackexchange",
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Prove $n < 2^n$ for all $n \geq 0$ using induction. Please verify for me?
Base case: $n = 0 $
$0 < 2^0$
$0 < 1$. This is true.
Inductive step: Suppose $n \geq 0$. Assume $P(k)$ is true if $k = n$. We must deduce that $P$ holds for $k+1$.
$n < 2^n$
$n - 2^n < 0$
$(n+1) - 2^{n+1}$
$n+1 - (2^n \times 2)$
$n+1 - (2^n + 2^n)$
$n - (2^n + 2^n) + 1$
$n - 2^n - 2^n + 1$
$(n - 2^n) - (2^n-1)$
We know $(n - 2^n) < 0$, that is to say $(n-2^n)$ is negative.
For all values of $n \geq 0, (2^n-1)$ is always a non-negative number.
A negative number minus a non-negative number is always negative, so
$(n - 2^n) - (2^n-1) < 0$ is true.
$(n - 2^n) < (2^n-1)$.
$n < (2^n + 2^n) -1$
$(n+1) < 2^{n+1}$
So $n < 2^n$ for all $n \geq 0$.
| There is an easier way. If $n<2^n$ for some $n$, then $$n+1<2^n+1 \leq 2^n+2^n=2^{n+1}$$.
| {
"language": "en",
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Linear Differential Equation but with piece wise function ( Cant Solve ) Find a continuous solution satisfying the DE
$$y' + y = f(x),$$
where
\begin{align}
f(x) &= \begin{cases}
1, & 0 \leq x \leq 1 \\
-1, & x>1\text{.}\end{cases}\\
y(0)&=1\end{align}
I am new to the forum so sorry guys for the format mistakes but really looking for the solution of this desperately & the 2nd question is
$$y' + 2xy = f(x),$$
where
\begin{align}
f(x) &= \begin{cases}
x, & 0 \leq x \leq 3 \\
0, & x >1\text{.}\end{cases}\\
y(0)&=2\end{align}
These questions are in my Book ODE by Zillls 3rd Edition Exercise 2.5 . Question number 57,58
Kindly help me out mates , i will be thankfull
| First equation
(Update Fixed, thanks to RodrigodeAzevedo)
Consider $0 \leq x \leq 1$, then:
$$\begin{cases}
y' + y = 1 \\
y(0) = 1
\end{cases} \Rightarrow y(x) = 1 \Rightarrow y(1) = 1.$$
Now, solve the differential equation for $x > 1$:
$$\begin{cases}
y' + y = -1 \\
y(1) = 1
\end{cases} \Rightarrow y(t) = 2e^{-x+1} - 1.$$
Finally:
$$y(x) = \begin{cases}1 & 0 \leq x \leq 1 \\
2e^{-x+1} - 1 & x > 1.
\end{cases}$$
Notice that for $x=1$, this function is continuous.
Second equation
This can be solved by using the very standard method of Separation of variables.
Consider $0 \leq x \leq 3$, then:
$$\begin{cases}
y' + 2xy = x \\
y(0) = 2
\end{cases} \Rightarrow \\
\frac{dy}{dx} = x(1-2y) \Rightarrow \\
\int_{y(0)}^{y(x)}\frac{dy}{1-2y} = \int_0^x s ds \Rightarrow \\
\int_{2}^{y(x)}\frac{dy}{1-2y} = \int_0^x s ds \Rightarrow \\
-\frac{1}{2}\left[\log\left(y(x)-\frac{1}{2}\right) - \log\left(\frac{3}{2}\right)\right] = \frac{1}{2}x^2 \Rightarrow \\
y(x) = \frac{3e^{-x^2}+1}{2} \Rightarrow
y(3) = \frac{3e^{-9}+1}{2}.$$
Now consider $x > 3$, then:
$$\begin{cases}
y' + 2xy = 0 \\
y(3) = \frac{3e^{-9}+1}{2}
\end{cases} \Rightarrow \\
\frac{dy}{dx} = -2xy \Rightarrow \\
\int_{y(3)}^{y(x)}\frac{dy}{y} = -2\int_0^x s ds \Rightarrow \\
\int_{\frac{3e^{-9}+1}{2}}^{y(x)}\frac{dy}{y} = -2\int_3^x s ds \Rightarrow \\
\log\left(y(x)\right) - \log\left(\frac{3e^{-9}+1}{2}\right) = -(x^2-9) \Rightarrow \\
y(x) = \frac{3e^{-9}+1}{2}e^{-(x^2-9)}$$
Again, this is continuous for $x=3$.
| {
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} |
What am I doing wrong in calculating the following limit? $$\lim_{x\to-2} \frac{x+2}{\sqrt{6+x}-2}=\lim_{x\to-2} \frac{1+2/x}{\sqrt{(6/x^2)+(1/x)}-2/x^2}$$ Dividing numerator and denominator by $x \neq0$
$$\frac{1+2/-2}{\sqrt{(6/4)+(1/-2)}-2/4}=\frac{0}{1/2}=0$$ but the limit is $4$ according to Wolfram Alpha?
| One may set $u=x+2$, then, as $x \to -2$, we have $u \to 0$, giving
$$
\frac{x+2}{\sqrt{6+x}-2}=\frac{u}{\sqrt{u+4}-2}\times\frac{\sqrt{u+4}+2}{\sqrt{u+4}+2}=\sqrt{u+4}+2 \to 4.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
How to prove that A and B are similar Let be $$A=\begin{pmatrix}
\frac{-3}{2} & 2 & \frac{-1}{2} \\
\frac{-1}{2} & 0 & \frac{1}{2} \\
\frac{1}{2} & -2 & \frac{3}{2}
\end{pmatrix},
B=\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}$$
Prove that A and B are similar.
I know if we can find a matrix $P$ so that $A=P^{-1}BP$ we have that they are similar, but I haven't found anywhere how to find such $P$. Can someone give me a hint?
| The easiest way I know would be to show that $A^2$ and $B^2$ are nonzero, but $A^3=B^3=0$. There is only one $3$-by-$3$ nilpotent matrix of nilpotency index $3$, up to conjugation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding the matrix for a linear transformation on a vector space when the basis changes Let B={$u_1,u_2,u_3$} as basis of Vector Space V, and Let T: V→V be the linear operator defined by,
$$
[T]_B=\begin{bmatrix}
-3 & 4 & 7 \\
1 & 0 & -2 \\
0 & 1 & 0 \\
\end{bmatrix}
$$
Find $[T]_{B'}$. B'={$v_1,v_2,v_3$} is basis of V defined by $v_1 = u_1, v_2 = u_1 + u_2, v_3 = u_1 + u_2 + u_3$.
I have solved questions by my effort, but I'm not sure for this answer. Please check and give me a hint for this question. Thank you.
My Answer:
$v_1 =\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}$, $v_2 =\begin{pmatrix}
1\\
1\\
0\\
\end{pmatrix}$, $v_3 =\begin{pmatrix}
1\\
1\\
1\\
\end{pmatrix}$
and, $T(v_1) =\begin{pmatrix}
-3 & 4 & 7 \\
1 & 0 & -2 \\
0 & 1 & 0 \\
\end{pmatrix}\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}=\begin{pmatrix}
-3\\
1\\
0\\
\end{pmatrix}$, $T(v_2) =\begin{pmatrix}
-3 & 4 & 7 \\
1 & 0 & -2 \\
0 & 1 & 0 \\
\end{pmatrix}\begin{pmatrix}
1\\
1\\
0\\
\end{pmatrix}=\begin{pmatrix}
1\\
1\\
1\\
\end{pmatrix}$$, T(v_3) =\begin{pmatrix}
-3 & 4 & 7 \\
1 & 0 & -2 \\
0 & 1 & 0 \\
\end{pmatrix}\begin{pmatrix}
1\\
1\\
1\\
\end{pmatrix}=\begin{pmatrix}
8\\
-1\\
1\\
\end{pmatrix}$
$[T(v_1)]_{B'}$ is consisted by $-4v_1+v_2$, $[T(v_2)]_{B'}$ is consisted by $v_3$, $[T(v_3)]_{B'}$ is consisted by $9v_1-2v_2+v_3$.
finally, vector representaion of $[T]_{B'} = \begin{pmatrix}
-4 & 0 & 9 \\
1 & 0 & -2 \\
0 & 1 & 1 \\
\end{pmatrix}$.
| Here is a trick.
Let $B' = \begin{pmatrix} 1&1&1\\0&1&1\\0&0&1\end{pmatrix}$ This is the basis of B' in terms of the basis of B.
To change the basis of T.
$[T]_B = [B'^{-1}TB']_{B'}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
build absolute value equations know solution We have absolute value equations with unknown coefficients:
$$|x + a| = b$$
and we know the solutions:
$$x = 11 \text{ and } x = 5$$
We need to find $a$ and $b$. From
$$11 + a = b \\
5 + a = -b$$
we get $a = -8$ and $b = 3$.
But we can try another way:
$$11 + a = -b \\
5 + a = b$$
and get $a =8$ and $b = -3$, which is not correct, apparently we need take absolute value from $b$.
How can I formulate this rule and explain it to another person?
| We want to solve $|x + a| = b$ where we know that the solutions for $x$ are 11 and 5. Since it's an absolute value, $b$ must be positive and therefore $b > -b$. We might try to solve
$$11 + a = -b \qquad 5 + a = b$$
But $11 + a > 5 + a$ and $-b$ cannot be greater than $b$. So we are left with
$$11 + a = b \qquad 5 + a = -b$$
which can be solved using simultaneous equations to get $a = -8$ and $b = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluating the inverse trigonometric limit $\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}$
$$
\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}
$$
I was doing some questions on limits, I saw one in which there is $\arccos x$. I am stuck there, not able to proceed.
Can you give me some hint?
| First of all, $$\arccos2x\sqrt{1-x^2}=\dfrac\pi2-\arcsin2x\sqrt{1-x^2}$$
Now using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $, $$2\arcsin x=\begin{cases}\arcsin2x\sqrt{1-x^2} &\mbox{if } |x|\le\dfrac1{\sqrt2} \\ \pi-\arcsin2x\sqrt{1-x^2} & \mbox{if } x>\dfrac1{\sqrt2}\\-\pi-\arcsin2x\sqrt{1-x^2} & \mbox{if } x<-\dfrac1{\sqrt2}
\end{cases}$$
$$\implies\lim_{x\to\frac1{\sqrt2}^+}\frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}=2\lim_{x\to\frac1{\sqrt2}^+}\frac{\arcsin x-\dfrac\pi4}{x-\frac{1}{\sqrt{2}}}=\dfrac{d(\arcsin x)}{dx}_{(\text{ at } x=\pi/4)}$$
and $$\lim_{x\to\frac1{\sqrt2}^-}\frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}=-2\lim_{x\to\frac1{\sqrt2}^-}\frac{\arcsin x-\dfrac\pi4}{x-\frac{1}{\sqrt{2}}}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Volume enclosed by $(x^2+y^2+z^2)^2=x$ I need to calculate the volume of solid enclosed by the surface $(x^2+y^2+z^2)^2=x$, using only spherical coordinates.
My attempt: by changing coordinates to spherical: $x=r\sin\phi\cos\theta~,~y=r\sin\phi\sin\theta~,~z=r\cos\phi$ we obtain the Jacobian $J=r^2\sin\phi$. When $\phi$ and $\theta$ are fixed, $r$ varies from $0$ to $\sqrt[3]{\sin\phi\cos\theta}$ (because $r^4=r\sin\phi\cos\theta$). Keeping $\theta$ fixed, we let $\phi$ vary from $0$ to $\pi$. Thus the volume equals:
$$V=\int\limits_{0}^{\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\sqrt[3]{\sin\phi\cos\theta}}r^2\sin\phi ~dr ~d\phi ~d\theta=0$$
Which is obviously wrong. What am I doing wrong?
| It's a solid of revolution. Let $h=\sqrt{y^{2}+z^{2}}$, then $(x^{2}+h^{2})^{2}=x$.
$\therefore \; h^{2}=\sqrt{x}-x^{2} \:$ where $\, 0\leq x \leq 1$.
\begin{align*}
V &= \int_{0}^{1} \pi h^{2} dx \\
&= \pi \int_{0}^{1} \left( \sqrt{x}-x^{2} \right) dx \\
&= \pi \left[ \frac{2}{3} x^{\frac{3}{2}}-
\frac{1}{3} x^{3} \right]_{0}^{1} \\
&= \frac{\pi}{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Using generating function to solve initial value problems I have a hard exam coming up and something I've struggled with since week 1 of semester is initial value problems. How would I go about solving:
(a) $u_{n} - 7u_{n-1} = 3 * 7^n : u_0 = 4 $
(b) $u_{n} - 4u_{n-1} + 4u_{n-1} = 3*2^n : u_0 = 6/u_1 =1 $
(c) $u_{n} - 3u_{n-1} = 2 * 5^n : u_0 = 7 $
Any help is appreciated.
| Part (a) and (c) follow the form:
$$f_{n} - a f_{n-1} = b \, c^{n}.$$
Since the generating function method is being asked then that is what will be demonstrated next.
\begin{align}
\sum_{n=0}^{\infty} \left( f_{n+1} - a f_{n} \right) t^{n} &= \frac{b}{1-c t} \\
\sum_{n=1}^{\infty} f_{n} t^{n-1} - a F(t) &= \\
\frac{1}{t} \, \left( F(t) - f_{0} \right) - a F(t) &= \\
(1 - a t) F(t) &= f_{0} + \frac{b}{1- c t}
\end{align}
which gives
$$F(t) = \sum_{n=0}^{\infty} f_{n} \, t^{n} = \frac{f_{0}}{1 - a \, t} + \frac{b}{(1 - a \, t)(1 - c \, t)}.$$
Now:
$$\frac{1}{(1 - a \, t)(1 - c \, t)} = \frac{1}{a - c} \, \left( \frac{a}{1 - a \, t} - \frac{c}{1 - c \, t} \right) $$
which leads to
\begin{align}
\sum_{n=0}^{\infty} f_{n} \, t^{n} &= \left( f_{0} + \frac{a b}{a-c} \right) \, \frac{1}{1 - a\, t} - \frac{b c}{a-c} \, \frac{1}{1 - b \, t} \\
&= \sum_{n=0}^{\infty} \left[ \left( f_{0} + \frac{a b}{a - c} \right) \, a^{n} - \frac{b c}{a-c} \, b^{n} \right] \, t^{n}.
\end{align}
This yields
$$f_{n} = \left( f_{0} + \frac{a b}{a - c} \right) \, a^{n} - \frac{b c}{a-c} \, b^{n}.$$
Applying the given values will provide the desired answers. Following a similar pattern will yield the answer to part (b).
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a,b>0$ and $a+b=1\;,$ Then minumum value of $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ is If $a,b>0$ and $a+b=1\;,$ Then minumum value of $\displaystyle \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2$ is
$\bf{My\; Try::}$ Let $a=\sin^2 \theta$ and $b=\cos^2 \theta\;,$ Then We have to minimize
$$\displaystyle f(\theta) = (\sin^2 \theta+\csc^2 \theta)^2+(\cos^2 \theta+\sec ^2 \theta) = \sin^4 \theta+\csc^4 \theta+\cos^4 \theta+\sec^4 \theta+4$$
So $$f(\theta) = 1-2\sin^2 \theta\cos^2 \theta+\frac{1}{\sin^2 \theta \cos^2 \theta}+4$$
So $$f(\theta) = 1-\frac{\sin^2 2\theta}{2}+\frac{4}{\sin^2 \theta}+4$$
Now Put $\sin^2 2\theta = t\;, t\in \left(0,1\right]$
So we get $$f(t)=5-\frac{t}{2}+\frac{4}{t}\;,$$ So we get $\displaystyle f'(t) = -\frac{1}{2}-\frac{4}{t^2}<0\;\forall t \in (0,1]$
So $$f(1)_{\min} = 5-\frac{1}{2}+4=9-\frac{1}{2}=\frac{17}{2}$$
But answer given as $$\frac{25}{2}$$
plz help me, Where I am wrong, Thanks
|
Where I am wrong
In the following part :
$$f(\theta) = 1-2\sin^2 \theta\cos^2 \theta+\frac{1}{\sin^2 \theta \cos^2 \theta}+4$$
This is not correct.
$$\begin{align}f(\theta)&=\sin^4\theta+\cos^4\theta+\frac{1}{\sin^4\theta}+\frac{1}{\cos^4\theta}+4\\&=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta+\frac{(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta}{\sin^4\theta\cos^4\theta}+4\\&=1-2\sin^2\theta\cos^2\theta+\frac{1-2\sin^2\theta\cos^2\theta}{\sin^4\theta\cos^4\theta}+4\end{align}$$
As you did, you can put $\sin^22\theta=t$ to see that $$f(t)=\frac{16}{t^2}-\frac t2-\frac 8t+5$$
is decreasing for $0\lt t\le 1$, so the answer is $f(1)=\frac{25}{2}$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given $\tan 3x=4$, find the value of $\tan^2 x+\tan ^2(120+x)+\tan^2 (60+x)$ Given $\tan 3x=4$, find the value of $S=\tan^2 x+\tan ^2(120+x)+\tan^2 (60+x)$
I expanded each of $\tan (120+x)$ and $\tan (60+x)$ getting as
$$S=\tan^2 x+\left(\frac{\tan 120+\tan x }{1-\tan 120 \tan x}\right)^2+\left(\frac{\tan 60+\tan x }{1-\tan 60 \tan x}\right)^2$$ so
$$S=\tan^2 x+\left(\frac{-\sqrt{3}+\tan x }{1+\sqrt{3} \tan x}\right)^2+\left(\frac{\sqrt{3}+\tan x }{1-\sqrt{3} \tan x}\right)^2$$
so
$$S=\tan^2 x+\frac{\left(4\tan x-\sqrt{3}(1+\tan^2x)\right)^2+\left(4\tan x+\sqrt{3}(1+\tan^2x)\right)^2}{(1-3\tan^2 x)^2}$$ so
$$S=\tan^2 x+\frac{\left(32 \tan^2 x+6(1+\tan^2 x)^2\right)}{(1-3\tan^2 x)^2}$$
but if i proceed further i dot think i will get an expression in terms of $\tan 3x$.
| As $a=\tan x,$
$b=\tan(60^\circ+x),\tan3(60^\circ+x)=\cdots=\tan3x,$
and $c=\tan(120^\circ+x),\tan3(120^\circ+x)=\cdots=\tan3x$
The roots of
$$4=\tan3y=\dfrac{3\tan y-\tan^3y}{1-3\tan^2y}$$
$$\iff\tan^3y-12\tan^2y-3\tan y+4=0$$ are $a,b,c$
We need $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Given $a_1,a_{100}, a_i=a_{i-1}a_{i+1}$, what's $a_1+a_2$? I've been given the following puzzle
Let $a_1, a_{100}$ be given real numbers. Let $a_i=a_{i-1}a_{i+1}$ for $2\leq i \leq 99$. Further suppose that the product of the first $50$ is $27$, and the product of all the $100$ numbers is also $27$.
Find $a_1+a_2$.
I tried the following, looking at the sequence for a moment we see:
$$
a_2=a_1\, a_3\\
a_3=a_2\,a_4\\
\vdots\\
a_{99}=a_{98}\,a_{100}
$$
So $$a_2=\frac 1 {a_{99}} \prod_i a_i =\frac {27}{a_{99}}$$
Looking at the other elements, we find that $$a_3=\frac {a_2} {a_1}, a_4=\frac {a_2}{a_1 a_2}, a_5=\frac {a_2}{a_1a_2a_3},\dots , a_n=\frac {a_2}{\prod_{i=1}^{n-2} a_i}$$
Then, $$27=\prod_i a_i=\prod_{1\leq i\leq 100} \frac {a_2}{\prod_{k=1}^{i-1}a_i}$$
The last product I believe is much more complicated than what I should have gotten...
Has anyone ran into this puzzle before?
| Note that since $a_i=a_{i-1}a_{i+1}$, then, for $i\geqslant 3$, we also have $a_{i-1}=a_{i-2}a_i$. Plug this in the first formula, we have $a_i=a_{i-2}a_ia_{i+1}\Rightarrow a_{i-2}a_{i+1}=1\Rightarrow a_{i+3}=\frac{1}{a_i}.$
Then, the sequence is in the form of $a_1=a,a_2=ab,a_3=b,a_4=\frac{1}{a},a_5=\frac{1}{ab},a_6=\frac{1}{b},a_7=a,\dots$.
$a_{i+6}=a_i$. So $a_{50}=ab$ and $a_{100}=\frac{1}{a}$. Hence, $1=\prod_{i=51}^{100}=\frac{b}{a}\Rightarrow a=b$. Also, $27=\prod_{i=1}^{50}=a\cdot (ab)=a^3\Rightarrow a=3\Rightarrow a_1+a_2=3+9=12$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof that $3^n | 2^{3^n} + 1$ Question:
Proof by induction that $3^n | 2^{3^n} + 1$.
Attempt: $$ 2^{3^{n+1}} + 1 = 2^{3^n} 2^3 + 1 = 2^{3^n} 2^3 + 1 + 2^3 - 2^3 =
2^3( 2^{3^n} + 1 ) + 1 -2^3$$
And the first is $3^n |$ but second I don't know how to proof that.
| $$2^{3^{n+1}}+1 = (2^{3^n})^3 + 1 = (2^{3^n}+1)((2^{3^n})^2 - 2^{3^n} + 1).$$ Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a+b+c=0$ what is the value of $\frac{a^2}{2a^2 +bc }+\frac{b^2}{2b^2 +ca }+\frac{c^2}{2c^2 +ab }$ Let $s=\frac{a^2}{2a^2 +bc }+\frac{b^2}{2b^2 +ca }+\frac{c^2}{2c^2 +ab }$.
If we use inequality $\frac{x^2}{a}+\frac{y^2}{b} \ge \frac{(x+y)^2}{(a+b)}$ we get $s \ge 0$ as $a+b+c=0$.
Again $s \le \frac{a^2}{bc }+\frac{b^2}{ca }+\frac{c^2}{ab }=\frac{a^3}{abc }+\frac{b^3}{abc }+\frac{c^3}{abc }=3$ as $a+b+c=0$ implies $a^3+b^3+c^3=3abc$.
Thus $0 \le s \le 3$. Putting any one of $a$,$b$,$c$=0 we get $s=1$. Hence my hunch is $s=1$, but i can't prove it. Can anyone help?
| Since for $a+b+c=0$ we have $a^3+b^3+c^3=3abc$, we obtain:
$$\sum_{cyc}\frac{a^2}{2a^2+bc}=\frac{\sum\limits_{cyc}a^2(2b^2+ac)(2c^2+ab)}{\prod\limits_{cyc}(2a^2+bc)}=\frac{\sum\limits_{cyc}(4a^2b^2c^2+4a^3b^3+a^4bc)}{\sum\limits_{cyc}(3a^2b^2c^2+4a^3b^3+2a^4bc)}=$$
$$=\frac{\sum\limits_{cyc}(4a^2b^2c^2+4a^3b^3+a^4bc)}{\sum\limits_{cyc}(3a^2b^2c^2+4a^3b^3+a^4bc+a^2b^2c^2)}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Finding the second derivative of$f(x)=x^2\sqrt{4-x}$
Find the second derivative of the function following:
$$f(x)=x^2\sqrt{4-x}$$
Here I go...
$$f(x)= x^2(4-x)^{1\over 2}$$
\begin{align*}
f'(x) &= 2x(4-x)^{1\over 2}+{1\over 2}x^2(4-x)^{-{1\over 2}}(-1)\\
&= 2x(4-x)^{1\over 2} - {1 \over 2}x^2(4-x)^{-{1\over 2}}\\
&={1\over 2}x(4-x)^{-{1\over 2}}[4(4-x)-x]\\
&= {1\over 2}x(4-x)^{-{1\over 2}}(16-5x)\\
f''(x)&= {-{1\over 4}}x(4-x)^{-{3\over 2}}(-1)({1\over 2})(16-5x)+(-5)[{1\over 2}x(4-x)^{-{1\over 2}}]\\
&= {-{1\over 8}}x(4-x)^{-{3\over 2}}(16-5x)-{5\over 2}(4-x)^{-{1\over 2}}\\
&= {-{1\over 8}}x(4-x)^{-{3\over 2}}[(16-5x)-20(4-x)]\\
&={-{1\over 8}}x(4-x)^{-{3\over 2}}(-64+15x^2)
\end{align*}
I think I messed up on the second derivative. Could anyone show me the steps for the second derivative? Thanks!
| Notice that
$$f'(x)=\frac12(4-x)^{-1/2}(16x-5x^2)$$
Then
\begin{align*}
f''(x)&=\frac14(4-x)^{-3/2}(16x-5x^2)+(4-x)^{-1/2}(8-5x)\\
&=\frac{4(4-x)(8-5x)+(16x-5x^2)}{4(4-x)^{3/2}}\\
&=\frac{15x^2-96x+128}{4(4-x)^{3/2}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Subtracting $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$ $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$
To solve the problem I first dissembled the equation on the denominator
$ \frac{(x+3)}{(x-1)*(x+1)} - \frac{(x-2)}{(x+1)^2}$
I multiplied the denominator together and to do this, I think I have to multiply the top part as well right? This is where i get confused, I forgot how to do this problem as its been a long time. How would i go on to solve this?
| The common factor can be taken to front as coefficient in the numerator as well as in the denominator.
$$ \frac{x+3}{(x-1)(x+1)} - \frac{x-2}{(x+1)^2}=\frac{1}{(x+1)} \cdot [\frac{x+3}{(x-1)} - \frac{x-2}{(x+1)}] $$
$$=\frac{1}{(x+1)} \cdot \frac{(x^2+4 x+ 3) - (x^2- 3 x + 2)} {(x^2-1)} $$
$$=\frac{1}{(x+1)} \frac{(7 x + 1)} {(x^2-1)}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$3$ digits numbers in which digits are repeated.
Total number of $3$ digit number which can be formed by using the digits $1,2,3,4,3,2,1$
$\bf{My\; Try::}$ Total no. of $3$ digits numbers in which exactly $2$ digits are identical, are
$112,113,114,221,223,224,331,332,334$
So Total numbers of $3$ digits numbers are $\displaystyle \frac{3!}{2!}+\frac{3!}{2!}+\frac{3!}{2!}+\frac{3!}{2!}+\frac{3!}{2!}+\frac{3!}{2!}+\frac{3!}{2!}+\frac{3!}{2!}+\frac{3!}{2!}$
So we get sum $ = 27$
And Total no. of $3$ digits numbers in which all digits are distinct,are
$123,124,234,341$
So Total numbers of $3$ digits numbers are $\displaystyle3!+3!+3!+3!=24$
So we get Total $ = 27+24=51.$
Is my solution is right, If not Then how can I calculate It, Thanks
| Yes, your solution is correct. In the second case, you want to find the number of 3 digit numbers with distinct digits where the digits are taken from $\{1,2,3,4\}$. This number is just $P(4,3)=4 \times 3 \times 2 = 24$ because the first digit can be chosen in 4 ways, the second in 3 ways (since the second digit can be any digit except the first digit), and the third digit can be chosen in 2 ways (since it can be any of the four digits available except the two digits chosen for the first and second positions).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Equilibrium in a system of nonlinear differential equations I have two questions about a specific system of differential equations. First, if a complex number can be an equilibrium point. Second, and related with the first question, how can I verify that $(0,0)$ is the only solution that satisfy $X'=0$ and $Y'=0.$ The system is as follows:
$X'=xy^3-xy^2$
$Y'=-y^3-3x^4$
Thanks a lot!
| We have two coupled polynomial ODEs
$$\dot x = xy^3 - xy^2 \qquad \qquad \dot y = -y^3 - 3x^4$$
To find the equilibria, solve the system of polynomial equations
$$0 = xy^3 - xy^2 \qquad \qquad 0 = -y^3 - 3x^4$$
Using SymPy,
>>> from sympy import roots, solve_poly_system
>>> x, y = symbols('x y')
>>> p1 = x * y**3 - x * y**2
>>> p2 = -y**3 - 3 * x**4
>>> solve_poly_system([p1, p2], x,y)
which outputs
\begin{equation*}\left [ \left ( 0, \quad 0\right ), \quad \left ( - \frac{\sqrt{2}}{6} 3^{\frac{3}{4}} - \frac{\sqrt{2} i}{6} 3^{\frac{3}{4}}, \quad 1\right ), \quad \left ( - \frac{\sqrt{2}}{6} 3^{\frac{3}{4}} + \frac{\sqrt{2} i}{6} 3^{\frac{3}{4}}, \quad 1\right ), \quad \left ( \frac{\sqrt{2}}{6} 3^{\frac{3}{4}} - \frac{\sqrt{2} i}{6} 3^{\frac{3}{4}}, \quad 1\right ), \quad \left ( \frac{\sqrt{2}}{6} 3^{\frac{3}{4}} + \frac{\sqrt{2} i}{6} 3^{\frac{3}{4}}, \quad 1\right )\right ]\end{equation*}
Hence, the origin is the only (real) equilibrium point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using AM-GM inequality to prove prove that $$x^4 + y^4 + z^4 \geq xyz(x+y+z)$$
This AM-GM inequalities are seriously stumping me. I'd appreciate a full proof and explanation and hints for proving other inequalities like this. Thanks.
| Use: $$a^2+b^2+c^2\ge ab+bc+ca$$
Proof:$$2(a^2+b^2+c^2)\ge 2(ab+bc+ca) \Leftrightarrow$$
$$\Leftrightarrow(a-b)^2+(b-c)^2+(c-a)^2\ge0$$
Then
$$x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xy^2z+x^2yz+xyz^2=$$
$$=xyz(x+y+z)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can someone show me how to simplify $\sin\left(\arccos\left(\frac{x}{x+1}\right)\right)$ I am needing help learning how to simplify the following equation:
$$\sin\left(\arccos\left(\frac{x}{x+1}\right)\right)$$
Also any steps on how to get the answer would be greatly appreciated!
| as Rahul said... let $\arccos(\frac{x}{x+1})=\theta$ so $\cos\theta=\frac{x}{x+1}$ but you want $\sin\theta$ which is $\sqrt{1-\cos^2\theta}$.
*
*Remember $\arccos t$ is between $0$ and $\pi$, so $\sin(\arccos t)$ is positive.
so the answer would be $$\sqrt{1-(\frac{x}{x+1})^2}=\sqrt{1-\frac{x^2}{(x+1)^2}}=\sqrt{\frac{(x+1)^2-x^2}{(x+1)^2}}=\sqrt{\frac{2x+1}{(x+1)^2}}=\frac{\sqrt{2x+1}}{x+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How to find the expected value of this game? How can I create a function that figures out the expected value for a value that can change.
Example:
I can flip a (fair) coin $n$ number of times.
The pot starts at $\$1$.
If I lose, $50$ cents is added to the pot.
If I win, I take what is in the pot and the pot is reset to $\$1$.
How can I figure out my expected value over $n$ number of coin flips?
The best I have come up with so far is:
$$\sum_{i=0}^n ((\text{currentPot} + (\text{incrmntOnLoss} \cdot \text{ numOfLosses})) \cdot \text{winChance})$$
but this doesn't take into account that the value can reset to $\$1$ after any flip.
It seems to be me that it would be some sort of recursive solution, but I'm unsure how that would be solved in math form
| Let $e_{n, p}$ be the expected value over $n$ coin flips if the initial value of the pot is $p$.
$e_{1,p}$ is easy to calculate; $$e_{1,p} = \frac 12 \cdot p = \frac p2$$
Now given the independence of different flips, we can write in general
$$e_{n,p} = \frac 12(p + e_{n-1, 1}) + \frac 12 e_{n-1, p+1/2}$$
Using this we write out the first few terms:
$$e_{2,p} = \frac 12 \cdot (p + e_{1,1}) + \frac 12 \cdot (e_{1,p+1/2}) = p/2 + 1/4 + p/4 + 1/8 = \frac 34p + \frac 38$$
$$e_{3,p} = \frac 12 (p + e_{2,1}) + \frac 12 e_{2,p + 1/2} = p/2 + 9/16 + 3/8p + 3/16 + 3/16 = \frac 78 p + \frac{15}{16}$$
$$e_{4,p} = \frac 12 (p + e_{3,1}) + \frac 12e_{3,p + 1/2} = p/2 + 7/16 + 15/32 + 7/16p + 7/32 + 15/32 =\frac{15}{16}p + \frac{51}{32}$$
This strongly suggests that the solution is in the form
$$e_{n,p} = \frac{2^n - 1}{2^n}p + \frac{\alpha_n}{2^{n+1}}$$
Who is $\alpha_n$ though? Since it looks too hard to guess, we just plug it in the general recurrence relation to find that $\alpha_n$ must satisfy
$$\alpha_n = 2\alpha_{n-1} + 2^n + 2^{n-1} - 3$$
The general solution is $\alpha_n = 2^{n-1}(c + 3n - 6) + 3$ and since $a_2 = 3$ we get $c = 0$. So in the end $\alpha_n = 2^{n-1}(3n - 6) + 3$ and our general solution is in the form
$$e_{n,p} = \frac{2^{n}-1}{2^n}p + \frac{2^{n-1}(3n - 6) + 3}{2^{n+1}}$$
You are interested in $p=1$, so the solution is
$$e_{n,1} = \frac{2^{n}-1}{2^n} + \frac{2^{n-1}(3n - 6) + 3}{2^{n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int_0^{\pi/2}\frac{\cos{x}}{2-\sin{2x}}dx$ How can I evaluate the following integral?
$$I=\int_0^{\pi/2}\frac{\cos{x}}{2-\sin{2x}}dx$$
I tried it with Wolfram Alpha, it gave me a numerical solution: $0.785398$.
Although I immediately know that it is equal to $\pi /4$, I fail to obtain the answer with pen and paper.
I tried to use substitution $u=\tan{x}$, but I failed because the upper limit of the integral is $\pi/2$ and $\tan{\pi/2}$ is undefined.
So how are we going to evaluate this integral? Thanks.
| There are 5 wonderful solutions already. I want to share with you one more alternative which is long but hopefully interesting.
Using $\sin 2x=2\sin x\cos x$ and multiplying both denominator and numerator by $1-\sin x \cos x$ rewrites$$
I=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos x+\sin x \cos ^2x}{1-\sin ^{2} x \cos ^{2} x} d x
$$
Splitting the numerator into 2 parts and letting respectively $\sin x \mapsto x$ and $\cos x\mapsto x $ yields
$$
\begin{array}{l} \displaystyle I=\frac{1}{2} \int_{0}^{1} \frac{1+x^{2}}{x^{4}-x^{2}+1} d x \\
\displaystyle \quad =\frac{1}{2} \int_{0}^{1} \frac{\frac{1}{x^{2}}+1}{x^{2}+\frac{1}{x^{2}}-1} d x \\
\displaystyle \quad =\frac{1}{2} \int_{0}^{1} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1} \\
\displaystyle \quad =\frac{1}{2}\left[\tan ^{-1}\left(x-\frac{1}{x}\right)\right]_{0}^{1} \\
\displaystyle \quad =\frac{1}{2}\left[0-\left(-\frac{\pi}{2}\right)\right] \\
\displaystyle \quad =\frac{\pi}{4}
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Calculate the area of a sphere drilled by two cylinders.
Let $S$ be the sphere given by the equation $x^2+y^2 +z^2 =4$ cut with
$z \geq 0$. Now, we drill the semisphere that is left with two
vertical cylinders of radius $1$, whose axes are respectively on the
points $(0,1,0)$ and $(0,-1,0)$. Calculate the area of the surface
that is left.
I know that the area of the semisphere of radius two is $8\pi$, but I don't know how to compute the area of the intersection between the cylinders and the semisphere. Any help with that would be highly appreciate.
Thanks in advance!
|
Parametrization of the surface cut out by one cylinder
The image above shows the intersection of the hemisphere with the cylinder whose axis passes through $(0,1,0)$ and whose equation is
$$
x^2+(y-1)^2=1
$$
The surface cut out by the cylinder can be parametrized with Cartesian coordinates as
$$
\vec\Sigma=\left(x,y,\sqrt{4-x^2-y^2}\right),\quad 0\le x^2+(y-1)^2\le1\\
\vec\Sigma_x\times\vec\Sigma_y=\begin{vmatrix}
\hat i&\hat j&\hat k\\
1& 0&-{x\over\sqrt{4-x^2-y^2}}\\
0& 1&-{y\over\sqrt{4-x^2-y^2}}\\
\end{vmatrix}\\
|\vec\Sigma_x\times\vec\Sigma_y|={2\over\sqrt{4-x^2-y^2}}$$
Calculation of surface area cut out by one cylinder
Area of the surface cut out by one cylinder is
$$\mathrm{
\int_0^{2}\int^{\sqrt{1-(y-1)^2}}_{-\sqrt{1-(y-1)^2}}{2\over\sqrt{4-x^2-y^2}}\,dx\,dy\\
=2\int_0^{2}\int^{\sqrt{2y-y^2}}_{-\sqrt{2y-y^2}}{1\over\sqrt{4-x^2-y^2}}\,dx\,dy\\
=4\int_0^{2}\int^{\sqrt{2y-y^2}}_{0}{1\over\sqrt{4-y^2-x^2}}\,dx\,dy\\
=4\int_0^2\arcsin{\sqrt{2y-y^2\over4-y^2}}\,dy\\
=4\int_0^2\arctan{\sqrt{2y-y^2\over4-2y}}\,dy\\
=4\int_0^2\arctan\sqrt{y\over2}\,dy\\
=16\int_0^1v\arctan v\,dv\quad\left(v=\sqrt{y\over2}\right)\\
=16\left[{v^2\over2}\arctan v\Big{|}_0^1-\int_0^1{v^2\over2(1+v^2)}dv\right]\\
=16\left[{\pi\over8}-{1\over2}\int_0^1\left(1-{1\over1+v^2}\right)dv\right]\\
=8\left[{\pi\over4}-\left(v-\arctan v\right)\Big{|}_0^1\right]\\=\color{blue}{4(\pi-2)}
}$$
Final answer
As $2$ identical holes have been drilled and the area of the hemisphere is ${1\over2}4\pi\times2^2$ sq units, the area of the surface that is left is
$$
{1\over2}4\pi\times2^2-2\times\color{blue}{4(\pi-2)}=16
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\cos A \cdot \cos 2 A \cdot \cos 4 A \cdots \cos 2^{n-1} A = \frac{\sin 2^n A}{2^n \sin A}$ Just a bit of background on the question:
When proving:
$$\cos\frac{\pi}{15}\cdot \cos\frac{2\pi}{15} \cdot \cos\frac{3\pi}{15}\cdot \cos\frac{4\pi}{15} \cdot \cos\frac{5\pi}{15} \cdot \cos\frac{6\pi}{15}\cdot \cos\frac{7\pi}{15} = \frac{1}{128}$$
The following formula was used:
$$\cos A \cdot \cos 2A \cdot \cos 2^2A \cdot \cos 2^3A\cdot \cdots \cdot \cos2^{n-1}A = \frac{\sin 2^n A}{2^n \sin A}$$
I'm just interested to see how this is derived. The question assumes that this formula is memorised and known beforehand, but it looked interesting to me since I'd actually never come across it before.
If anyone's interested, I'll link full proof for the question here
Cheers in advance!
| Turn the double-angle formula for sine "inside out". Put in
$$\begin{align}\cos(A)& =\frac{\sin(2A)}{2·\sin(A)} \\[6pt]
\cos(2A)&=\frac{\sin(4A)}{2·\sin(2A)}
\end{align}$$
etc., and use telescoping.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
System of diophantine equations $x^2+3y=u^2$, $y^2+3x=v^2$
Solve the following system of Diophantine equations(the unknowns are positive integers):
$$
\left\{
\begin{array}{c}
x^2+3y=u^2 \\
y^2+3x=v^2
\end{array}
\right.
$$
I worked as follows:
subtract the two equations to get: $4x^2-4y^2-12(x-y)=9y^2-9x^2\ \implies\ ... (x-y)(13x+13y-12)=0\implies x=y\ or\ 13x+13y-12=0$
The first equation has infinite answers and the second has none(since $gcd(13,13)$ does not divide $12$), am I right??
| $$x^2-y^2-3(x-y)=u^2-v^2$$
$$(x-y)(x+y-3)=(u+v)(u-v)$$
Assuming $u^2-v^2\neq 0$, we have:
$$\dfrac{x-y}{u-v}=\dfrac{u+v}{x+y-3}=\dfrac{r}{s}$$ where $\gcd(r,s)=1$.
$$x-y=\dfrac{r(u-v)}{s}$$
$$x+y=\dfrac{s(u+v)}{r}+3$$
since $x,y$ are integers, and $u,v$ have the same parity, then there exist $p,q$ such that: $$u-v=ps $$ $$ u+v=qr $$
Hence,
$$x-y=pr $$
$$x+y=qs+3 $$
with $$x=\dfrac{qs+pr+3}{2}$$ $$y=\dfrac{qs-pr+3}{2}$$
$$u=\dfrac{qr+ps}{2} $$ $$v=\dfrac{qr-ps}{2} $$ where either $qr$ and $ps$ have the same parity and $pr$ and $qs$ have opposite parity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Linear application Let $f:\mathbb{R}^3\to\mathbb{R}^3$ be a linear application and let $\{e_1,e_2,e_3\}$ the canonical basis of $\mathbb{R}^3$.
We know that $\operatorname{Im} f=\langle(1,1,3), (0,1,1)\rangle$ and that
$$2f(e_2)-f(e_3)=e_1-e_2+e_3$$
Which is the matrix associated to $f$ with respect the canonical basis?
In order to have the matrix I should know $f(e_1), f(e_2), f(e_3)$ but I don't know how to do it.
| Assume that
$$fe_1=\begin{pmatrix}1\\1\\3\end{pmatrix}\;\;,\;\;\;fe_2=\begin{pmatrix}0\\1\\1\end{pmatrix}\;\;,\;\;fe_3=\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}$$
Observe that we have already
$$\text{Im}\,f\subset\left\langle\;\begin{pmatrix}1\\1\\3\end{pmatrix}\;,\;\;\begin{pmatrix}0\\1\\1\end{pmatrix}\;\right\rangle$$
Now, we're given
$$e_1-e_2+e_3=\begin{pmatrix}1\\\!-1\\1\end{pmatrix}=2fe_2-fe_3=\begin{pmatrix}\!-u_1\\2-u_2\\2-u_3\end{pmatrix}\;\implies\;fe_3=\begin{pmatrix}\!-1\\3\\1\end{pmatrix}$$
and observe that
$$fe_3\in\left\langle\;\begin{pmatrix}1\\1\\3\end{pmatrix}\;,\;\;\begin{pmatrix}0\\1\\1\end{pmatrix}\;\right\rangle\implies\text{Im}\,f=\left\langle\;\begin{pmatrix}1\\1\\3\end{pmatrix}\;,\;\;\begin{pmatrix}0\\1\\1\end{pmatrix}\;\right\rangle\implies$$
$$[f]_e=\begin{pmatrix}1&0&\!-1\\1&1&3\\3&1&1\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of special tridiagonal determinant Let $A_{n}$ be the following tridiagonal determinant of order $n:$
\begin{vmatrix}
a_{0}+a_{1}& a_{1}& 0& 0& \cdots& 0& \quad0\\
a_{1}& a_{1}+a_{2}& a_{2}& 0& \cdots& 0& \quad0\\
0& a_{2}& a_{2}+a_{3}& a_{3}& \cdots& 0& \quad0\\
\vdots& \vdots& \vdots& \vdots& & \vdots& \quad\vdots\\
0& 0& 0& 0& \cdots& a_{n-1}& \quad a_{n-1}+a_{n}
\end{vmatrix}
Find the value of $A_{n}.$
As we know,$$A_{n}=(a_{n-1}+a_{n})A_{n-1}-a^{2}_{n-1}A_{n-2}\Longrightarrow$$ $$\begin{bmatrix}
A_{n}\\
A_{n-1}
\end{bmatrix}=\begin{bmatrix}
a_{n-1}+a_{n}& -a^{2}_{n-1}\\
1& 0
\end{bmatrix}\begin{bmatrix}
A_{n-1}\\
A_{n-2}
\end{bmatrix}=$$$$\begin{bmatrix}
a_{n-1}+a_{n}& -a^{2}_{n-1}\\
1& 0
\end{bmatrix}\begin{bmatrix}
a_{n-2}+a_{n-1}& -a^{2}_{n-2}\\
1& 0
\end{bmatrix}\cdots\begin{bmatrix}
a_{2}+a_{3}& -a^{2}_{2}\\
1& 0
\end{bmatrix}\begin{bmatrix}
A_{2}\\
A_{1}
\end{bmatrix}.$$
But it is not easy to deal with
$$\begin{bmatrix}
a_{n-1}+a_{n}& -a^{2}_{n-1}\\
1& 0
\end{bmatrix}\begin{bmatrix}
a_{n-2}+a_{n-1}& -a^{2}_{n-2}\\
1& 0
\end{bmatrix}\cdots\begin{bmatrix}
a_{2}+a_{3}& -a^{2}_{2}\\
1& 0
\end{bmatrix}$$
| Here is an observation that may save some labor later on. At first glance, the above recurrence relation only makes sense for $n>2$. But for $n=2$, the recurrence requires $A_2=(a_2+a_1)A_1-a_{1}^2 A_0$; since But $A_2=(a_0+a_1)(a_1+a_2)-a_1^2$ and $A_1=(a_1+a_0)$, the recurrence will be valid at $n=2$ if $A_0=1$. Similarly, for $n=1$ the recurrence would give $A_1 = (a_1+a_0)A_0-a_{-1}^2 A_{-1}$ which is valid if $A_{-1}=0$. Introducing $T_n=\begin{bmatrix} a_{n-1}+a_n & -a_{n-1}^2 \\ 1 & 0 \end{bmatrix}$ for convenience, one then has
$$\begin{bmatrix} A_n \\ A_{n-1} \end{bmatrix}
=T_n
\begin{bmatrix} A_{n-1} \\ A_{n-2} \end{bmatrix}
=\cdots =e_1^T (T_n T_{n-1}\cdots T_1)\begin{bmatrix} A_{0} \\ A_{-1} \end{bmatrix}=(T_n T_{n-1}\cdots T_1)e_1.$$ So obtaining $A_n$ amounts to finding the upper-left element of $T_n T_{n-1}\cdots T_1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A BBP-type series The BBP-type series
$$ \frac{\pi}{2} \, \left( \frac{\alpha^{2}}{5} \right)^{\frac{1}{4}} = \sum_{n=0}^{\infty} \left[ \frac{1}{10 n + 1} + \frac{\alpha}{10 n + 3} - \frac{\alpha}{10 n + 7} - \frac{1}{10 n + 9} \right],$$
with golden ratio $\alpha = \frac{1 + \sqrt{5}}{2}$ can be obtained by a particular Sine series. The questions proposed here are:
*
*Can multiple Fourier-Sine/Cosine series yield the same result?
*Are there non-Fourier series methods which yield this series?
| Hint. One may recall the following series representation of the digamma function,
$$
\sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{u+k}
\right)=\psi(u+1)+\gamma, \qquad u >-1,
$$ giving
$$
\sum_{k=0}^{\infty} \left( \frac{1}{k+a} - \frac{1}{k+b}
\right)=\psi(b)-\psi(a),\qquad a>0,\,b>0. \tag1
$$
From $(1)$ one gets
$$
\begin{align}
&10\cdot\sum _{n=0}^{\infty } \left(\frac{1}{10 n+a}+\frac{\alpha }{10 n+b}-\frac{\alpha }{10 n+c}-\frac{1}{10 n+d}\right)
\\\\&=-\psi\left(\frac{a}{10}\right)-\alpha\: \psi\left(\frac{b}{10}\right)+\alpha\: \psi\left(\frac{c}{10}\right)+\psi\left(\frac{d}{10}\right)
\end{align}
$$ then, putting
$$
a=1,\quad b=3,\quad c=7,\quad d=9,
$$ using Gauss' digamma theorem one deduces
$$
\begin{align}
10\cdot\sum_{n=0}^{\infty} \left( \frac{1}{10 n + 1} + \frac{\alpha}{10 n + 3} - \frac{\alpha}{10 n + 7} - \frac{1}{10 n + 9} \right)
=\left(\sqrt{5+2\sqrt{5}}+\sqrt{5-2\sqrt{5}}\:\alpha\right)\pi
\end{align}
$$
from which one obtains the announced result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Without calculating $A^4$ prove that $A^4\in Span\{A,I\}.$ Let
$$A=
\begin{bmatrix}
-1 & 6 & -9 \\
-11 & 24 & -33 \\
-6 & 12 & -16 \\
\end{bmatrix}
$$
a) Without calculating $A^4$ prove that $A^4\in Span\{A,I\}.$
b) Write $A^n$ in a form of $a_nA+b_nI$
If matrix $A^4\in Span\{A,I\}$ then $A^4=\alpha A+ \beta B$.
I had an idea to find eigenvalues and eigenvectors of $A$, so I can diagonalize the matrix and say that $A^4=SD^4S^{-1}$. However, I am not sure is that done without calculating $A^4$.
Thank you all in advance.
| The characteristic polynomial of $A$ is
$$ p_A(\lambda) = \det(\lambda I - A) = \lambda^3 - 7\lambda^2 + 16\lambda - 12 = (\lambda - 3)(\lambda - 2)^2. $$
Since part (b) asks you to show in particular that $A^2 \in \operatorname{span} \{ I, A \}$, the minimal polynomial of $A$ must be of degree at most two. Since the minimal polynomial of $A$ must have the same linear factors as the characteristic polynomial, we then must have $m_A(\lambda) = (\lambda - 3)(\lambda - 2) = \lambda^2 - 5\lambda + 6$. This indeed holds:
$$ (A - 3I)(A - 2I) = \begin{pmatrix} -4 & 6 & -9 \\ -11 & 21 & -33 \\ -6 & 12 & -19 \end{pmatrix} \begin{pmatrix} -3 & 6 & -9 \\ -11 & 22 & -33 \\ -6 & 12 & -18 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$
(you can deduce this without calculating by noticing that $1 = \operatorname{rank}(A - 2I) = 3 - \dim \ker(A - 2I)$ and so $A$ is diagonalizable and the minimal polynomial must be $(\lambda - 3)(\lambda - 2)$).
Thus, $A^2 = 5A - 6I$ and at this point you know that $A^n \in \operatorname{span} \{ I, A \}$ for all $n \geq 0$ (and in particular $n = 4$) without explicitly calculating $A^4$ or the coefficients. Finally, if we write $A^n = a_n A + b_n I$ we must have
$$ A^{n + 1} = a_{n+1} A + b_{n+1}I = A(A^n) = A(a_n A + b_n I) = a_n A^2 + b_n A = a_n (5A - 6I) + b_n A = (5a_n + b_n) A - (6a_n) I. $$
Since $A,I$ are linearly independent we get $a_{n+1} = 5a_n + b_n, b_{n+1} = -6a_n$ which is a recursion relation that can be solved explicitly together with the initial conditions $a_1 = 5, b_1 = -6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality problem, for positive $a,b,c$, if $abc=1$, then $\frac{1}{1+a+b^2}+\frac{1}{1+b+c^2}+\frac{1}{1+c+a^2}\leq1$ I need help or guidance in solving this inequality that I am battling for 3 days now. I have tried everything that comes to mind, but I am stuck. The inequality is as $$\sum_\textrm{cyc}\frac{1}{1+a+b^2}\leq1,\ abc=1, \ a,b,c>0$$ This can be written as $$\frac{1}{1+a+b^2}+\frac{1}{1+b+c^2}+\frac{1}{1+c+a^2}\leq1$$
I have tried Jensen's on the function $\frac{1}{1+x}$, AM-GM on the denominator, Cauchy-Schwarz for each term but I keep getting stuck.
| The following technique can deal with many of the above type of inequalities. For example, Problem 3 of IMO 2005 can be solved using the same idea (but a little bit easier). People participating in math Olympiads should keep it in mind.
For any $k$, applying Cauchy-Schwarz inequality we have
$$(1+a+b^2)(c^{2k} + a^{2k-1} + b^{2k-2}) \ge (c^k+a^k+b^k)^2,$$
yielding $$\frac{1}{1+a+b^2} \le \frac{c^{2k} + a^{2k-1} + b^{2k-2}}{(a^k+b^k+c^k)^2}.$$
Similarly for the other two terms, taking the sum, it remains to prove
$$\sum (a^{2k} + a^{2k-1} + a^{2k-2}) \le (a^k+b^k+c^k)^2$$
or equivalenty
$$(a^{2k-1} + b^{2k-1} + c^{2k-1}) + (a^{2k-2} + b^{2k-2} + c^{2k-2}) \le 2(a^kb^k+b^kc^k+c^ka^k).$$
Re-writing this in homogeneous form:
$$(abc)^{1/3}(a^{2k-1} + b^{2k-1} + c^{2k-1}) + (abc)^{2/3}(a^{2k-2} + b^{2k-2} + c^{2k-2}) \le 2(a^kb^k+b^kc^k+c^ka^k) \quad (*)$$
It suffices to find a value of $k$ for which this inequality holds for any $a,b,c > 0$. In this case, $k=2/3$ is such a value. Indeed, if $k=2/3$, denote $x=a^{1/3},y=b^{1/3},z=c^{1/3}$, $(*)$ becomes
$$xyz(x+y+z) + x^2y^2z^2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right) \le 2(x^2y^2+y^2z^2+z^2x^2),$$
or equivalently $$xyz(x+y+z) \le x^2y^2+y^2z^2+z^2x^2,$$
which is just $rs+st+tr \le r^2+s^2+t^2$ with $r=xy,s=yz,t=zx$.
Remark.
*
*The inequality $(*)$ is not true for the trivial value $k=1$. (This was indeed my first try.)
*I could have not mentioned $k$ at all, just implicitly set it to $2/3$, and put $x=a^{1/3},y=b^{1/3},z=c^{1/3}$ at the beginning of the proof. However, I think it's useful to show how I obtained the solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Triple Integration problem, stuck. I have a doubt over this triple integral: I have to integrate
$$\iiint x\ \text{d}x$$
between
$$z = x^2 + y^2 ~~~~~~~ and ~~~~~~~ 4x + z = -2$$
But I feel like dumb because I cannot even start!
It's between $z = x^2 + y^2$ and $z = -4x - 2$.
If I equal them I get
$$x^2 + y^2 + 4x + 2 = 0$$
Which I cannot solve, if it makes sense.
Cylindrical Try
$$x = R\cos\theta ~~~~~ y = R\sin\theta$$
Then it is
$$z = R^2$$
$$4R\cos\theta + R^2 = -2$$
Which means
$$R = -2\cos\theta \pm \sqrt{2}\sqrt{2\cos^2\theta + 1}$$
Right?
OR
Without cylindrical coordinates, I can solve for $y$ and $x$.
$$y^2 = z - x^2$$
and using $z = -4x -2$ I find
$$y = \pm \sqrt{2 - (x+2)^2}$$
The same for $x$ finding
$$x = -2+\sqrt{2} ~~~~~ x = -2-\sqrt{2}$$
Then I have limits, right??
| $$I=\int_{-2-\sqrt{2}}^{-2+\sqrt{2}}\int_{-\sqrt{2-(x+2)^2}}^{\sqrt{2-(x+2)^2}}\int_{-2-4x}^{x^2+y^2}x\,dzdydx$$
let $x=-2+r\cos \theta$ and $y=r\sin\theta$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $x^2 + y^2 + z^2 \ge 35$ if $x+3y+5z \ge 35.$ Show that $x^2 + y^2 + z^2 \ge 35$ if $x+3y+5z \ge 35.$
I have tried everything (proof by contradiction, etc.) but I can't seem to get it. The book didn't give any constraints whatsoever. Any hints would be appreciated. Thank you.
| I suggest a solution for who know Lagrange multiplier.
Let $f(x,y,z)=x^2+y^2+z^2$, and $g(x,y,z)=x+3y+5z=a\ge 35$. Then
\begin{align}
\nabla f(x,y,z)&=(2x,2y,2z)\\
\nabla g(x,y,z)&=(1,3,5)\ne \mathbf{0}
\end{align}
Let $\lambda$ be a real number such that $\nabla f = \lambda \nabla g$, then
$$
2x=\lambda,\quad 2y=3\lambda,\quad 2z=5\lambda.
$$
Thus
$$
x+3y+5z=\frac{\lambda}{2}+\frac{9\lambda}{2}+\frac{25\lambda}{2}=\frac{35\lambda}{2},
$$
so $\lambda=\frac{2a}{35}$. Therefore,
\begin{align}
x^2+y^2+z^2&=\frac{\lambda^2}{4}+\frac{9\lambda^2}{4}+\frac{25\lambda^2}{4}\\
&= \frac{35}{4}\cdot \lambda^2\\
&=\frac{35}{4}\cdot \frac{4a^2}{35^2}\\
&=\frac{a^2}{35}\\
&\ge 35.
\end{align}
| {
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Does this equation have no solutions? The question is this :
The source from where I got this question was devoid of any answers to it, so I came here, this is how I proceeded :
LHS :
$((((({(x)^x})^{2x})^{3x})^{....x^2})^2 = (((((x)^{2x^2})^{3x})^{....x^2})^2 =...........= (x^{(x^x)x!})^2 = x^{2(x^x)x!} $
RHS :
$\sqrt{x^2\sqrt{(x-1)x\sqrt{(x-2)x\sqrt{...........2x\sqrt{x\sqrt x}}}}} = \sqrt{x\sqrt{(x-1)\sqrt{(x-2)\sqrt{...........2\sqrt{1\sqrt 1}}}}}*x^{(1-\frac{1}{2^{x+1}})} $ $= x^{1/2}*(x-1)^{1/4}*(x-2)^{1/8}*....... *{2}^{(\frac{1}{2^{x-1}})}* x^{(1-\frac{1}{2^{x+1}})} $
Now I thought of taking $\log$ of both RHS and LHS from which I could deduce
LHS:
$\log x^{2(x^x)x!} = 2(x^x)x!\log x = 2*x^{x+1}(x-1)!\log x $
RHS:
$\log (x^{1/2}*(x-1)^{1/4}*(x-2)^{1/8}*....... *{2}^{(\frac{1}{2^{x-1}})}* x^{(1-\frac{1}{2^{x+1}})})$ $= \frac{\log x}{2}+\frac{\log (x-1)}{4}+ \frac{\log (x-2)}{8}+..........+ \frac{\log 2}{2^{x-1}}+ {(1-\frac{1}{2^{x+1}})} \log x $
Now equating LHS = RHS I get :
$ 2*x^{x+1}(x-1)!\log x = \frac{\log x}{2}+\frac{\log (x-1)}{4}+ \frac{\log (x-2)}{8}+..........+ \frac{\log 2}{2^{x-1}}+ {(1-\frac{1}{2^{x+1}})} \log x$ $\implies x^{x+1}(x-1)! =\frac{ \frac{\log x}{2}+\frac{\log (x-1)}{4} + \frac{\log (x-2)}{8}+..........+\frac{\log 2}{2^{x-1}}+ {(1-\frac{1}{2^{x+1}})} \log x}{2\log x} $
now in the RHS of above equation I only found $\log x$ in 3 places : denominator, first place of numerator and in the last place of numerator; I assumed terms from $\frac {\log (x-1)}{4*2\log x}$ to $\frac{\log 2}{2^{x-1}*2\log x}$ were becoming too small to take into calculation, so final equation could be written down to :
$ x^{x+1}(x-1)! = \frac{ \frac{\log x}{2} + {(1-\frac{1}{2^{x+1}})} \log x}{2\log x} =\frac{\frac{1}{2} +1- \frac{1}{ 2^{x+1} }}{2} = \frac{3}{4} - \frac{1}{ 2^{x+2} } $$\implies x^{x+1}(x-1)! = \frac{3}{4} - \frac{1}{ 2^{x+2} }$
the above equation is where I am forced to stop, please guide me after that ? Or did I take a wrong approach from start itself ?
(If you can, do tag it with appropriate tags; I could not find the suitable ones for this problem)
| The equation can be written as:
$x^{2x^{x} x!}=\sqrt{x^{\frac{ 1}{x! x^{x}} }}$.
By squared up the two members, we get:
$x^{4x^{x} x!}=x^{\frac{ 1}{x! x^{x}}}$.
Since this is the equality between two powers, for them to be equal they must have the same basis and the same exponent; therefore, since the basics are the same, we can write:
$ 4x^{x} x!= \frac{ 1}{x! x^{x}}$.
Solving for $x^{x} x!$, you get two values $±\frac{1}{2}$.
Noting that the expression $(x-1)!x^{x+1}$, is none other than $x^{x} x!$, as
$(x-1)!x^{x+1}=(x-1)! x x^{x}=x!x^{x}$,
the solution is the positive real value $(x-1)!x^{x+1}=\frac{1}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Which number is greater, $11^{11}$ or $9^{12}$?
Which number is greater than $11^{11}$ or $9^{12}$?
My work so far:
$11^{11}=285311670611>9^{12}=282429536481$.
But to verify the validity of equality should be in the range of easily verifiable calculations.
| More generally, you are looking to prove: $x^x > (x-2)^{x+1}$. This can be done by taking log both sides, and its easier. Consider $f(x) = x\ln x - (x+1)\ln (x-2)$ on $(11, \infty)$, and taking log we have: $f'(x) = \ln x + 1 - \ln(x-2) - \dfrac{x+1}{x-2}$. We have $f''(x) = \dfrac{1}{x} - \dfrac{1}{x-2}+ \dfrac{3}{(x-2)^2}= \dfrac{(x-2)^2-x(x-2) + 3}{x(x-2)^2}= \dfrac{7-2x}{x(x-2)^2} < 0\implies f'(x) > f'(\infty) = 0 \implies f(x) > f(11) $
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $
| I would multiply by the conjugate of denominator $$a=\frac{3 \sqrt{3}-4}{7-2 \sqrt{3}}=\frac{1}{37} \left(13 \sqrt{3}-10\right)$$ $$b=\frac{3 \sqrt{3}-8}{1-2 \sqrt{3}}=\frac{1}{11} \left(13 \sqrt{3}-10\right)$$ this makes $a<b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solutions of $\sin^2\theta = \frac{x^2+y^2}{2xy} $ If $x$ and $y$ are real, then the equation
$$\sin^2\theta = \frac{x^2+y^2}{2xy}$$
has a solution:
*
*for all $x$ and $y$
*for no $x$ and $y$
*only when $x \neq y \neq 0$
*only when $x = y \neq 0$
| First you need $x \ne 0$ and $y \ne 0$ in order for your fraction $\dfrac{x^2+y^2}{2xy}$ to make sense.
Then as $\sin \in [-1,1]$ you need $x^2+y^2 \leq 2xy$ which means:
$x^2 -2xy+y^2 \leq 0$ ie $(x-y)^2 \leq 0$
Therefore you have solutions on $\theta$ only if $x = y \ne 0$, and those solutions are $\theta = \dfrac{\pi}{2} \pmod \pi$
| {
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Inequality with square root $x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}}$ Good morning to everyone! The inequality is the following:$$ x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} $$. I don't know how to solve it. Here's what I tried: $$x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} \rightarrow 2x^2-11x+9+\sqrt{x^2-10x+9}\left(2x+1\right)\ge \:\sqrt{x^2-10x+9}$$ Therefore $ 2x+1 \ge 0 => x\ge \frac{1}{2} $ so $x$ belongs to $(\frac{1}{2},\infty) $ and $ x^2-10x+9 \ge 0 => x$ belongs to $ (-\infty,1] $ and $[9,\infty) $. Therefore the statements are contradicting each other.
| You have got yourself into a tangle by being too clever too soon, and squaring. It is better to simplify first. With careful simplification you don't need calculus or anything advanced at all.
To pick up a theme of Kenny Lau, put $u=x+\sqrt{x^2-10x+9}$. Then your inequality becomes $u\ge\sqrt{u}$, which is equivalent (following Gerry Myerson's comment) to $u\ge{1}$.
Translating back to $x$ language, this makes your inequality equivalent to $$x+\sqrt{x^2-10x+9}\ge{1}$$
Since the expression under the square root is equivalent to $(x-1)(x-9)$, we can write this as $$(x-1)+\sqrt{(x-1)(x-9)}\ge{0}$$ and putting $y=x-1$, we get $$y+\sqrt{y(y-8)}\ge{0}$$.
For the expression under the square root to make sense we need either $y\le{0}$ or $y\ge{8}$.
*
*$y\ge{8}$ certainly satisfies the inequality. Put into $x$ terms, this means $x\ge{9}$.
What about $y\le{0}$? Let's do another substitution, of $z=-y$, which means that $z\ge{0}$. This gives us $$-z+\sqrt{z(z+8)}\ge{0}$$
Dividing by $\sqrt{z}$ and rearranging, we get $$\sqrt{z+8}\ge\sqrt{z}$$
… which is always true.
*
*Therefore $z\ge{0}$ also satisfies the equality, which means $y\le{0}$, which means $x\le{1}$.
| {
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How to show that this function $f(x,y,) <0$? I would like to show that the function $f(x,y) = -5 x^4 + 4 y^2 - 5 y^4 + x^2 (4 - 6 y^2)$ is less than $0$ for $1 < x^2 + y^2 <2$ (at least). Upon rearranging, I obtain
$f(x,y) = -5(x^2+y^2)^2 + 4(x^2 +y^2 + x^2y^2)$
But then I don't know how to proceed to get $f(x,y)<0?
Thanks!
| Let us write this expression under the form:
$$f(x,y) = -5(x^2+y^2)^2 + 4(x^2 +y^2) + 4 x^2y^2$$
Let us convert this expression into polar coordinates. One gets:
$$F(r,\theta) = -5r^4 + 4r^2 + r^4 \sin ^2(2 \theta)$$
(using relationship $\sin (2 \theta) = 2 \sin \theta \cos \theta$).
Therefore:
$$F(r,\theta) = r^2 (r^2 (\sin ^2(2 \theta) - 5) + 4)$$
For $r>0$, the sign of $F$ is that of $g(r,\theta):=r^2 (\sin^2(2 \theta) - 5) + 4$
Let $1<r^2<2 \ \ \ (1) $. Taking into consideration that
$0-5 \leq \sin^2(2 \theta) - 5 \leq -4 \ \ \ (2)$
Multiplication of (1) by (2) gives:
$$-5 \leq r^2 (\sin^2(2 \theta) - 5) \leq -4$$
Adding 4 to the three parts gives the result: $-1 \leq g(r,\theta) \leq 0$.
| {
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Proving that $\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+\cdots+\frac{1}{(n+3)(n+4)}=\frac{n}{4(n+4)}$ by induction I've proved the base case where $n=1$ and made the assumption that $n=k$ is true, but I'm stuck on the $n=k+1$ part. I just cannot seem to get the algebra to work in my favor.
Here is the original:
$\displaystyle\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+…+\frac{1}{(n+3)(n+4)}=\frac{n}{4(n+4)}\qquad \forall\, n \in \mathbb N$
I get it to the following form and just run in circles:
$\displaystyle\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+...+\frac{1}{(k+3)(k+4)} +\frac{1}{(k+4)(k+5)}=\frac{k+1}{4(k+1+4)}$
Simplifying
$\displaystyle\frac{k}{4(k+4)} +\frac{1}{(k+4)(k+5)}=\frac{k+1}{4(k+1+4)}$
Is this in the correct form?
| This should help:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{(i+3)(i+4)}&= \sum_{i=1}^k\frac{1}{(i+3)(i+4)}+\frac{1}{(k+4)(k+5)}\\[1em]
&= \frac{k}{4(k+4)}+\frac{1}{(k+4)(k+5)}\\[1em]
&= \frac{k(k+5)+4}{4(k+4)(k+5)}\\[1em]
&= \frac{(k+1)(k+4)}{4(k+4)(k+5)}\\[1em]
&= \frac{k+1}{4(k+5)}.
\end{align}
See if you can spot where the inductive hypothesis is used and where some basic algebraic manipulation plays an important role.
| {
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Is $(2,5)$ the only solution? Find all pairs $(m,n)\in{\mathbb{N^2}}$ such that
$$(m^2-1)^3-n^2=2$$
Is $(2,5)$ the only solution?
| Maybe this helps
$(m^2-1)^3-n^2 =2\iff (m^2-1)^3-27=n^2-25$
$\iff (m^2-4)( (m^4-2m^2+1)+3(m^2-1)+9)=(n-5)(n+5)$
$\iff(m+2)(m-2)(m^4+m^2+7)=(n-5)(n+5)$
| {
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The value of $x^2+y^2+z^2+w^2$ Let$x,y,z,w$ satisfy
$$\frac{x^2}{2^2 - 1^2} +\frac{y^2}{2^2 - 3^2} +\frac{z^2}{2^2 - 5^2} +\frac{w^2}{2^2 - 7^2} =1$$
$$\frac{x^2}{4^2 - 1^2} +\frac{y^2}{4^2 - 3^2} +\frac{z^2}{4^2 - 5^2} +\frac{w^2}{4^2 - 7^2} =1$$
$$\frac{x^2}{6^2 - 1^2} +\frac{y^2}{6^2 - 3^2} +\frac{z^2}{6^2 - 5^2} +\frac{w^2}{6^2 - 7^2} =1$$
$$\frac{x^2}{8^2 - 1^2} +\frac{y^2}{8^2 - 3^2} +\frac{z^2}{8^2 - 5^2} +\frac{w^2}{8^2 - 7^2} =1$$
My work
$$\frac{x^2}{t - 1^2} +\frac{y^2}{t - 3^2} +\frac{z^2}{t - 5^2} +\frac{w^2}{t - 7^2} =1$$
where $t $ satisfy $4,16,36,64$
$$f(t)=0$$
$$f(t) = (t – 1)(t – 9)(t – 25)(t – 49)–x^2(t – 9)(t – 25)(t – 49) –y^2(t – 1)(t – 25)(t – 49) – z^2(t–1)(t–9)(t–49) – w^2(t–1)(t–9)(t–25)$$
then I compared the coefficient with different value of $t$ .
I want to know that is there any easier alternative methods for this .
| Expanding
\begin{equation*}
\frac{x^2}{t - 1^2} +\frac{y^2}{t - 3^2} +\frac{z^2}{t - 5^2} +\frac{w^2}{t - 7^2} =1
\end{equation*}
we get
\begin{equation*}
(t-1^2)(t-3^2)(t-5^2)(t-7^2) - (t-3^2)(t-5^2)(t-7^2)x^2 - \text{ similar terms } = 0
\end{equation*}
This biquadratic in $t$ has four roots $2^2, 4^2, 6^2, 8^2$ and coefficient of $t^3$ is $-(2^2+4^2+6^2+8^2)$. The coefficient of $t^3$ is also given by
\begin{equation*}
-(1^2+3^2+5^2+7^2) - (x^2+y^2+z^2+w^2)
\end{equation*}
and hence
\begin{equation*}
-(1^2+3^2+5^2+7^2) - (x^2+y^2+z^2+w^2) = -(2^2+4^2+6^2+8^2)
\end{equation*}
Hence we have
\begin{equation*}
x^2+y^2+z^2+w^2 = (2^2+4^2+6^2+8^2)-(1^2+3^2+5^2+7^2)=36
\end{equation*}
| {
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"url": "https://math.stackexchange.com/questions/1846993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the number of solutions to $ \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345$
Find the number of solutions of the equation
$$\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor =12345,$$
where $\lfloor\,\cdot\,\rfloor$ represents the floor function.
My work:
I use the fact that
$$\lfloor nx \rfloor =\sum_{k=0}^{n-1} \left\lfloor x +\frac kn \right\rfloor.$$
So the equation becomes
$$\lfloor x \rfloor +\sum_{k=0}^{1} \left\lfloor x +\frac k2 \right\rfloor +\sum_{k=0}^{3} \left\lfloor x +\frac k4 \right\rfloor +\sum_{k=0}^{7} \left\lfloor x +\frac k8 \right\rfloor \\
\qquad {}+\sum_{k=0}^{15} \left\lfloor x +\frac k{16} \right\rfloor +\sum_{k=0}^{31} \left\lfloor x+\frac k{32} \right\rfloor \\
= \lfloor x \rfloor + \left\lfloor x+\frac 12 \right\rfloor + \left\lfloor x+\frac 64 \right\rfloor + \left\lfloor x+\frac{28}{8} \right\rfloor \\
\qquad {}+ \left\lfloor x+\frac{120}{16} \right\rfloor + \left\lfloor x+\frac{496}{32} \right\rfloor$$
What should I do next?
| Your work, while a reasonable approach, does not lead to the answer. A slightly simpler approach does:
Let $p \in \Bbb{Z} = \lfloor x \rfloor$ and let $x= p +q$ with $0\leq q < 1$.
Then the left hand side is
$$L = 63p + \lfloor q\rfloor + \lfloor 2q\rfloor+ \lfloor 4q\rfloor+ \lfloor 8q\rfloor+ \lfloor 16q\rfloor+ \lfloor 32q\rfloor \geq 63p \\
L = 12345 \implies 12345 \geq 63p \implies p \leq 12345/63 < 196
$$
$$L = 63p + \lfloor q\rfloor + \lfloor 2q\rfloor+ \lfloor 4q\rfloor+ \lfloor 8q\rfloor+ \lfloor 16q\rfloor+ \lfloor 32q\rfloor \leq 63p + 1+3+7+15+31 \\
L = 12345 \implies 12345 \leq 63p + 57 \implies p \geq 12288/63 > 195
$$
So if a solution exists, then $p$ is an integer between about $195.05$ and $195.95$. No such integer exists, the the number of solutions is zero.
| {
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Prove by induction that $n^5-5n^3+4n$ is divisible by 120 for all n starting from 3 I've tried expanding $(n+1)^5-5(n+1)^3+4(n+1)$ but I end up with $120k+5(n^4+2n^3-n^2-2n)$ where k is any positive whole number, and I can't manipulate $5(n^4+2n^3-n^2-2n)$ to factor with 120.
| Using repeated differences and Newton's interpolation formula we get
$$
n^5-5n^3+4n = 120 \binom{n}{3} + 240 \binom{n}{4} + 120 \binom{n}{5}
$$
Although this identity suffices for answering the question, it also implies the simpler identity below:
$$
n ^5-5n^3+4n = 120 \binom{n+2}{5}
$$
which gives a crystal clear answer to the question.
If you must use induction, then:
\begin{align}
f(n+1)-f(n)
&=5 n^4+10 n^3-5 n^2-10 n\\
&= 5 (n+2) (n+1) n (n-1)\\
&= 5 (4!) \binom{n+2}{4}\\
&= 120 \binom{n+2}{4}
\end{align}
| {
"language": "en",
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"source": "stackexchange",
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Factoring out a $7$ from $3^{35}-5$? Please Note: My main concern now is how to factor $7$ from $3^{35}-5$ using Algebraic techniques, not how to solve the problem itself; the motivation is just for background.
Motivation:
I was trying to solve the following problem
What is the remainder when $10^{35}$ is divided by $7$?
I used the binomial formula: $\dfrac {(7+3)^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot 3^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot + 35 \cdot 3^{34} \cdot 7}{7} + \dfrac {3^{35}}{7}$
Therefore, $10^{35}$ will have the same remainder as $3^{35}$ when divided by $7$.
I was stuck here, and I wanted to try to reverse engineer the answer. I know the remainder is $5$. Therefore I should be able to write $3^{35} -5 + 5$ as $7k+5$. However, I have no idea how to factor out a $7$ from $3^{35}-5$, or from $10^{35}-5.$ How could I find this $k$ non-explicitly?
| If you insist on algebraic techniques then note $\ 3(3^{35}-5)\, =\, (3^{36}-1)-14$
and $\,\ 7 = \color{#c00}{3^2\!-\!3\!+\!1}\mid 3^6\!-\!1\mid 3^{36}-1\ $ so $\,7\,$ divides both summands above, so $\,7\mid 3^{35}\!-\!5$
where $\ \color{#c00}{x^2-x+1}\mid x^6-1\ $ since it divides the $\,\rm\color{#c00}{difference\ of\ squares}\,$ below
$$ x^6-1\ =\, (x^2-1)\,(\!\overbrace{x^4+x^2+1}^{\Large\color{#c00}{ (x^2+1)^2-x^2}}\!)$$
Remark $\ $ "Algebraic" factors that arise arise from polynomial factorizations are heavily employed when factoring integers of the form $\rm\:b^n\pm 1.\:$ A good place to learn about such is Wagstaff's splendid introduction to the Cunningham Project, whose goal is to factor numbers of the form $\rm\:b^n\pm 1.\:$ There you will find mentioned not only old results such as Legendre (primitive divisors of $\rm\:b^n\pm 1\:$ are $\rm\,\equiv 1\pmod{2n},$ but also newer results, e.g. those exploiting cyclotomic factorizations. e.g. see below.
Often number identities are more perceptively viewed as special cases of function or polynomial identities. For example, Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations (e.g. see below).
$$\begin{array}{rl}
x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\[0.5em]
\dfrac{x^6 + 3^2}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\[0.5em]
\dfrac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\[0.5em]
\dfrac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36)
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Polynomial ,divides and Induction Proof?
$\text{The polynomial } x-y \;\text{divides the polynomial}\; x^2-y^2 \text{ and } x^3-y^3 \text{because}\; x^2-y^2 = (x+y)(x-y) \text{ and } x^3-y^3=(x-y)(x^2+xy+y^2.) \; \text{for every natural number n } \quad x-y \;\text{ divides }\; x^n-y^n \text{ prove by induction.}$
What I am confused about in this example is when one makes $n=k$. Which variable is made equal to k.Where is $n$. Is it fair to assume that what the actual problem wants you to infer this. $x^n +y^n = x^n+y^n$
(i) Basis Step: $x^n-y^n = x^n-y^n$
$x-y = x-y$
$0=0$
(ii) Inductive step : $x^k-y^k = x^k-y^k$
WNTS: RHS: $n =k+1$
$x^{k+1} + y^{k+1}$
LHS: $x^{k+1} + y^{k+1} +x^k-y^k $
Is this a fair assumption to make I feel as if their is something fundamentally missing from my logic. Any suggestions would be good.
| For the inductive step, suppose $x-y$ divides $x^n-y^n$ for some $n$. You have to prove that, under this hypothesis, $x-y$ divides $x^{n+1}-y^{n+1}$.
Hint:
Rewrite it as $$x^{n+1}-y^{n+1}=x(x^n-y^n)+xy^n - y^{n+1}$$
and make partial factorisations.
Some details:
By the induction hypothesis, $x^n-y^n=(x-y)q(x,y)$, so
$$x(x^n-y^n)+xy^n - y^{n+1}=(x-y)q(x,y)+ (x-y)y^n=(x-y)\bigl(q(x,y)+y^n\bigr).$$
Actually, the induction step defines a recurrence relation for the quotient of $x^n-y^n$ by $x-y$, which allows to prove the complete factorisation formula.
| {
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The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$ what is the value of $P\times Q\times R$ Problem
The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$
And we are asked to find $P\times Q\times R$ by using the identities:
$P(x)=Q(x)\times D(x)+R(x)$
where $P(x)$ is the equations and $Q(x)$ is the quotient and $D(x)$ is the divisor and $R(x)$ is the remainder.
My thought process
so I had set $$ax^2+bx+c=Q(x)\times(x-6)\times(x-P)+0$$
and in the next equation:
$$cx^2+bx+a=Q(x)\times(x-P)(x-Q)+0$$
I really don't know where to go from here
| Considering roots of $ax^2+bx+c$ as $\frac{-b\pm \delta}{2a}$ the roots of $cx^2+bx+a$ will be $\frac{-b\pm \delta}{2c}$. So, the product of all the four roots are $$6PQR=\frac{(-b+ \delta)(-b-\delta)}{2a}\times\frac{(-b+ \delta)(-b-\delta)}{2c}=\frac{4ac}{4ac}=1$$ Thus, we have $$PQR=\frac16$$
| {
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"url": "https://math.stackexchange.com/questions/1853465",
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"source": "stackexchange",
"question_score": "3",
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On the solution of the equation $z+ \alpha \left| z-1\right| + 2i = 0$ Question:-
Find the range of real number $\alpha$ for which the equation $z+ \alpha \left| z-1\right| + 2i = 0$; $z=x+iy$ has a solution. Also find the solution.
Attempt at a solution:-
On substituting $z=x+iy$ in the equation $z+ \alpha \left| z-1\right| + 2i = 0$, we get that the imaginary part of the complex variable $z$ should be $-2$, i.e $y=-2$. So, the complex variable is of the form $z=x-2i$.
As, the real part of the equation is also $0$, so we get the following equation for the real part
$$\begin{equation}x=-\alpha\sqrt{x^2-2x+5} \end{equation} \tag{1}$$
After squaring both sides of the equation, we arrive at the following quadratic equation.
$$\begin{equation} (\alpha^2-1)x^2-2\alpha^2x+5\alpha^2=0 \end{equation} \tag{2}$$
Now, as $x$ has to be real so $D \ge 0 \implies 5-4\alpha^2 \ge 0 \implies -\dfrac{\sqrt{5}}{2}\le \alpha \le \dfrac{\sqrt{5}}{2}$
Now, let's consider different cases.
Case 1:-
When $\alpha^2=1$, then from $(2)$, we get $x=\dfrac{5}{2}$. Now, after plugging $x=\dfrac{5}{2}$ in $(1)$, we get $\dfrac{5}{2}=-\alpha\left(\dfrac{5}{2}\right)$, so from this we conclude that when $x=\dfrac{5}{2}$, then, $\alpha=-1$
The place where I am getting stuck:-
I am not able to think up of the different solutions of $x$, when $\alpha^2 \neq 1$. If anyone can help me think of how to go about solving for the remaining cases.
| From $(1)$, $\alpha=0$ works. If $\alpha\not=0$, then
$$\sqrt{(x-1)^2+4}=\frac{x}{-\alpha}\gt 0\tag3$$
From $(2)$, for $-\frac{\sqrt 5}{2}\le \alpha\le\frac{\sqrt 5}{2}$ with $\alpha^2\not=1$,
$$x=\frac{\alpha^2\pm \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}$$
Case 1 : For $\alpha=0$, $x=0$.
Case 2 : For $\alpha^2=1$, you've already done. ($\alpha=-1,x=5/2$)
Case 3 : For $-\frac{\sqrt 5}{2}\le \alpha\lt -1$, from $(3)$, $x\gt 0$, so $x=\frac{\alpha^2\color{red}{\pm} \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}$.
Case 4 : For $-1\lt \alpha\lt 0$, from $(3)$, $x\gt 0$, so $x=\frac{\alpha^2\color{red}{+} \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}$.
Case 5 : For $0\lt \alpha\lt 1$, from $(3)$, $x\lt 0$, so $x=\frac{\alpha^2\color{red}{+} \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}$.
Case 6 : For $1\lt \alpha\le\frac{\sqrt 5}{2}$, from $(3)$, $x\lt 0$, so there is no solution.
Therefore, the range of real number $\alpha$ for which the equation $z+ \alpha \left| z-1\right| + 2i = 0$; $z=x+iy$ has a solution is $\color{red}{-\frac{\sqrt 5}{2}\le\alpha\lt 1}$.
And the solution is
$$\color{red}{x=\frac{\alpha^2\pm \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}\qquad\text{for $\ -\frac{\sqrt 5}{2}\le\alpha\lt -1$}}$$
$$\color{red}{x=\frac 52\qquad\text{for $\ \alpha=-1$}}$$
$$\color{red}{x=\frac{\alpha^2+ \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}\qquad\text{for $\ -1\lt\alpha\lt 1$}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the roots $x^4-4x^3-x^2-8x+4=0$ (contest math) So the problem is :
$x^4-4x^3-x^2-8x+4=0$, find all solutions
A tip that I have gotten, is to divide both sides by $x^2$. I've tried so, but I do not manage to see any further. Do anyone know how this tip could help me?
(Yes, I'm aware that the polynomial above can be factorized into two degree 2 polynomials, which promptly gives me the answer. But that factorization would be extremely hard to spot, which is why I'm asking about the dividing)
Thanks in advance :)
Edit: meant to write $x^2$, not $2$
| Divide $x^4-4x^3-x^2-8x+4=0$ with $x^2$ in order to get the following $x^2-4x-1-\frac{8}{x}+\frac{4}{x^2}=(x+\frac{2}{x})^2-4(x+\frac{2}{x})-5=(x+\frac{2}{x}-5)(x+\frac{2}{x}+1)$. Finally, result is $(x^2-5x+2)(x^2+x+2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$1+2+3+45+6+78+9=144$ what are other combinations Note that $$1+2+3+45+6+78+9 = 144$$ In how many other ways is it possible to make a total of $144$ using only $1, 2, 3, 4, 5, 6, 7, 8,$ and $9$ in that order and addition signs?
Sorry I am only in high school so dont over complicate the explanation. Thank you
| If we "bunch up" $8$ and $9$ to get $89$, the rest of the numbers must add up to $55$. Bunching $6$ and $7, 5$ and $6$, and $4$ and $5$ will make the sum too high. If we bunch $3$ and $4$ to get $34$, so we have $34+5+6+7+89=141$ which needs $3$ more.
So $$1+2+34+5+6+7+89=144$$.
That's the only possible $89$ bunch up as $12+3+4+5+6+7+89$ doesn't work.
You have a solution with a $78$ bunch up. Any other $78$ bunch up has $78+9=88$ requires the rest of the numbers to add to $56$. We can't bunch $56$. If we bunch up $45$ we get $45+6+78+9=138$ and we need $6$ more so we get your answer if we don't bunch up $45$ and bunch up $34$ we have $34+5+6+78+9=132$ and we need $12$ so we can do $12+34+5+6+78+9$.
If we bunch up $67$ we get $67+8+9=84$ and we need $60$ more. If we bunch $45$ we need $15$ more from $1,2,3$ so we can have $12+3+45+67+8+9$.
If we bunch $67$ but not $45$ we have $5+67+8+9=89$ so we need to get $51$ from $1,2,3,4$. The most we can do is $12+34$ so there is no other with $67$ bunch.
If we bunch up $56$ we have $56+7+8+9=80$ and we need $64$ more. If we bunch up the $34$ we need $30$ more which can't be done with only the $1$ and $2$.
If we don't bunch the $56$ we have $5+6+7+8+9=35$ and we need to get $109$ from $1,2,3,4$. The most we can get is $12+34=46$ so we can't go any further. Those four are the only solutions.
| {
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Evaluate $\int \frac{\sqrt{64x^2-256}}{x}\,dx$ QUESTION
Evaluate $$\int \frac{\sqrt{64x^2-256}}{x}\,dx$$
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
MY ATTEMPT
*
*Typed
$\newcommand{\dd}{\; \mathrm{d}}\int \frac{\sqrt{64x^2-256}}x \dd x \to
\int \frac{\sqrt{64(x^2-4)}}x \dd x \to
\int \frac{8\sqrt{x^2-4}}x \dd x$
Use $x=a\sec\theta$, $\dd x=a\sec\theta \tan\theta \dd \theta$.
$a=2$ $\to$ $x=2\sec\theta$, $\dd x=2\sec\theta \tan\theta \dd \theta$.
$=\int \frac{8\sqrt{4\sec^2\theta-4}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to
\int \frac{8\sqrt{4(\sec^2\theta-1)}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta $
$=\int \frac{8\sqrt{4\tan^2\theta}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to
\int \frac{8(2\tan\theta)}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta$
$=\int 16\tan^2\theta \dd \theta \to
16\int\tan^2\theta \dd \theta \to
\underset{\text{trig. formula}}{\underbrace{16(\theta+\tan\theta)+C}}$
$\Rightarrow 16(\tan\theta-\theta)+C = 16\tan\theta-16\theta$
$x=2\sec\theta$, $\sec\theta= \frac x2$
$\boxed{16\tan\left(\frac{\sqrt{x^2-4}}2\right) -16\sec^{-1}\left(\frac x2\right)+C}$
*
*Handwritten
| Since I am almost blind, I have a lot of problems reading the image.
Consider $$I=\int \frac{\sqrt{64 x^2-256}}{x}\,dx$$ What you apparently did is $x=2\sec(t)$, $dx=2 \tan (t) \sec (t)$ which make $$I=\int \tan (t) \sqrt{256 \sec ^2(t)-256}\,dt=16\int \tan (t) \sqrt{\tan ^2(t)}\,dt=16\int \tan^2 (t) \,dt$$ $$I=16\int (1+\tan^2(t)-1)\,dt=16 (\tan (t)-t)$$
| {
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Show that $\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}n+\mathcal{O}\left(\frac{1}{n^{3/2}}\right)$
How can i prove that $$\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}{n} +\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\tag{$*$}$$
using the following method :
note that :
$(1+x)^{\alpha}=1+\alpha x+\mathcal{O}(x^{2})\quad ( x\to 0) $
\begin{align}
\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}&=\frac{(-1)^n}{n} \left(1+\frac{(-1)^n\sqrt{n+1}}{n} \right)^{-1}\\
&\text{since } \sqrt{n+1}\sim \sqrt{n} \text{ then } \lim_{n\to+\infty}\frac{\sqrt{n+1}}{n}=0 \\
&\sim \frac{(-1)^n}{n}\left(1+\frac{(-1)^n\sqrt{n}}{n} \right)^{-1}\\
&= \frac{(-1)^{n}}n\left(1+\mathcal{O}\left( \frac{(-1)^n\sqrt{n}}{n}\right) \right)\\
&= \frac{(-1)^n}{n}+\mathcal{O}\left( \dfrac{(-1)^{2n}\sqrt{n}}{n^2}\right) \\
&=\frac{(-1)^n}{n}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)
\end{align}
*
*AM i right ?
| One may write, as $n \to \infty$,
$$
\begin{align}
\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}&=\frac{(-1)^n}{n} \left(1+\frac{(-1)^n\sqrt{n+1}}{n} \right)^{-1}\\
&= \frac{(-1)^n}{n}\left(1+\frac{(-1)^n}{\sqrt{n}}\cdot\sqrt{1+\frac1n}\: \right)^{-1}\\
&= \frac{(-1)^n}{n}\left(1+\frac{(-1)^n}{\sqrt{n}}\left(1+\frac1{2n}+\mathcal{O}\left(\dfrac{1}{n^2}\right)\right)\: \right)^{-1}\\
&= \frac{(-1)^n}{n}\left(1+\frac{(-1)^n}{\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\: \right)^{-1}\\
&= \frac{(-1)^n}{n}\left(1-\frac{(-1)^n}{\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\: \right)\\
&=\frac{(-1)^n}{n}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)
\end{align}
$$ as wanted.
| {
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"source": "stackexchange",
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Show that $(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\tfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\tfrac{1}{n^{3/2}} \right)$
I would like to show that :
$$\fbox{$(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)$}$$
by starting from the left side and get the right side
My Proof:
\begin{align*}
(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)&=(-1)^{n}\left(\dfrac{1}{\sqrt{n+1}+\sqrt{n} }\right)\\
&=(-1)^{n}\left(\sqrt{n+1}+\sqrt{n} \right)^{-1}\\
&=(-1)^{n}\left(\sqrt{n}\left(1+\sqrt{1+\dfrac{1}{n}}\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{\sqrt{n}}\left(1+\left(1+\dfrac{1}{n}\right)^{\dfrac{1}{2}} \right)^{-1}\\
&=\mbox{Note that :}\left(1+x\right)^{\alpha}=1+\mathcal{O}\left( x\right) \\
&=\dfrac{(-1)^{n}}{\sqrt{n}}\left(1+\left(1+\mathcal{O}\left(\dfrac{1}{n} \right)\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{\sqrt{n}}\left(2+\mathcal{O}\left(\dfrac{1}{n}\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\dfrac{1}{n}\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\mathcal{O}\left(\dfrac{1}{n}\right)\right) \right)\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\dfrac{1}{n}\right) \right)\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)\\
\end{align*}
$$\fbox{$(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)$}$$
*
*Am i right ?
| You van do it also by purely algebraic means:
$$\begin{align}
\frac{1}{\sqrt{n+1}+\sqrt{n}}&=\frac{1}{2\sqrt{n}}+\frac{\sqrt{n}-\sqrt{n+1}}{2\sqrt{n}(\sqrt{n+1}+\sqrt{n})}\\
&=\frac{1}{2\sqrt{n}}-\frac{1}{2\sqrt{n}(\sqrt{n+1}+\sqrt{n})^2}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Linear combination issue I have 4 vectors:
$u_1 = \begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix} $, $\; u_2 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{pmatrix} $, $\; u_3 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{pmatrix} $, $\; u_4 = \begin{pmatrix}
0 \\
1 \\
0 \\
\end{pmatrix} $
and I wanna express the following vector in terms of them:
$\; v = \begin{pmatrix}
2 \\
3 \\
4 \\
\end{pmatrix} $
I'm working this way:
First put vectors in a matrix and then put in rref:
$ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
1&1&1&1&3\\
1&0&0&0&4
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&0&0&1&1\\
0&-1&-1&0&2
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&-1&-1&0&2\\
0&0&0&1&1
\end{array}
\right] $ =>
$ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&1&1&0&-2\\
0&0&0&1&1
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&1&1&0&-2\\
0&0&0&1&1
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&0&0&0&4\\
0&1&1&0&-2\\
0&0&0&1&1
\end{array}
\right] $
but this result (4 -2 1) is meaningless to me (because $4u_1 -2u_2 + u_3 \ne v$)... does it make any sense? Can it represent some sort of coeficients of the linear combination?
If I swap $u_3$ for $u_4$ in the matrix, the (4 -2 1) is the same but it DOES make sense, because $4u_1 -2u_2 + u_4 = v$
How can I write v as combination of u's ? If I did the right way, how does this make sense?
| If I rewrite your augmented matrix as follows:
$\begin{bmatrix}
1&0&0&0&\\
0&1&1&0&\\
0&0&0&1&\\
\end{bmatrix}$
$\begin{bmatrix}
a_1\\
a_2\\
a_3\\
a_4\\
\end{bmatrix}$
$= \begin{bmatrix}
4\\
-2\\
1\\
\end{bmatrix}$
Where $a_1, \cdots a_4$ are the coefficients of $u_1,\cdots u_4$
$\begin{bmatrix}
a_1\\
a_2+a_3\\
a_4\\
\end{bmatrix}$
$= \begin{bmatrix}
4\\
-2\\
1\\
\end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Differentiate and simplify. $m(x) = \frac{x}{\sqrt{4x-3}}$ My work so far is:
\begin{align}
m'(x) &= \frac{(1)(\sqrt{4x-3})-(x)(1/2)(4x-3)^{-1/2}(4)}{(\sqrt{4x-3})^2} \\
&= \frac{\sqrt{4x-3} - 2x(4x-3)^{1/2}}{4x-3}
\end{align}
and now I'm stuck on how to simplify further
| In general we have $\frac{\partial}{\partial x} \frac{f(x)}{g(x)}= \frac{\frac{\partial f(x)}{\partial x} \times g(x) - \frac{\partial g(x)}{\partial x} \times f(x)}{g^2(x)}$. So in the case of your question, set $f(x)= x$ and $g(x) = (4x - 3)^{\frac{1}{2}}$. And we have:
$\frac{\partial}{\partial x} (x) = 1$ and
$\frac{\partial}{\partial x} ((4x - 3)^{\frac{1}{2}}) = \frac{1}{2} \times 4 \times (4x - 3)^{- \frac{1}{2}}= 2 \times (4x - 3)^{- \frac{1}{2}}$. So we have:
$\frac{\partial}{\partial x} \frac{x}{(4x - 3)^{\frac{1}{2}}}= \frac{1 \times (4x - 3)^{\frac{1}{2}} - 2 \times (4x - 3)^{- \frac{1}{2}} \times x }{4x - 3} = \frac{2x-3}{(4x - 3)^{\frac{3}{2}}}$.
And we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Remainder when $5^{5555}$ is divided by $10000$. Find the remainder when $5^{5555}$ is divided by $10000$.
A step by step guide with explanation for a beginner student in modular arithmetic is needed.
| Start by factoring $10000 = 10^4 = 2^4 \times 5^4$.
So we want to know the results mod $2^4$ and mod $5^4$.
The second is easiest: $5555 > 4$ so $5^{5555} \equiv 0 \mod 5^4$.
Now mod $2^4 = 16$: $\gcd(2^4, 5) = 1$ and $\phi(16) = 8$, so $5^8 \equiv 1 \mod 2^4$. $5555 \equiv 3 \mod 8$, so $5^{5555} \equiv 5^3 \equiv 13 \mod 2^4$.
You want a number $x \in [0,1,\ldots, 9999]$ with $x \equiv 0 \mod 5^4$ and $x \equiv 13 \mod 2^4$. The $0$ makes it especially easy. You'll have
$x = 5^4 y$, and $5^4 \equiv 1 \mod 2^4$, so $y \equiv x \equiv 13
\mod 2^4$. You can take $y = 13$ and thus $x = 8125$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to decompose $x^3-1$ I need to decompose $x^3-1$, I know the Binomial theorem, and finding roots of a polynomial, how should I approach this?
| We can just use binomial theorem:
Observe
$$(x-1)^3=x^3-3x^2+3x-1$$
So $$\begin{align}x^3-1&=(x-1)^3+3x^2-3x\\
&=(x-1)^3+3x(x-1)\\
&=(x-1)((x-1)^2+3x)\\
&=(x-1)(x^2+x+1)\end{align}$$
Hope this helps.
P.S.
If factoring $x^2+x+1$ is needed, recall we have a formula for quadratic polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Problem 1, Ch. 6 in Piskunov's, Differential and Integral calculus
Find the curvature of the curve at indicated points
$b^2x^2+a^2y^2=a^2b^2$ at $(0,b)$ and $(a,0)$
My attempt
$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$
Differentiating the implicit equation with respect to $x$,
$2b^2x+2a^2yy'=0\\
y'=-\frac{b^2}{a^2}{\cdot}\frac{x}{y}$
Differentiating again with respect to x,
$y''=-\frac{b^2}{a^2}{\cdot}\frac{y-xy'}{y^2}$
a) At $(0,b)$
$y'=0\\
y''=-\frac{b^2}{a^2}{\cdot}\frac{b-(0)(0)}{b^2}=-\frac{b}{a^2}$
$\displaystyle{\kappa=\frac{|-\frac{b}{a^2}|}{\left[1+\left(0\right)^2\right]^\frac{3}{2}}=\frac{b}{a^2}}$
b) At $(a,0)$
$y'=\infty\\
y''=-\frac{b^2}{a^2}{\cdot}\frac{\frac{y}{y'}-x}{\frac{y^2}{y'}}=\infty$
I am not sure, if my solution to part (b) is correct. The book says, the answer must be $\frac{a}{b^2}$.
| we get the curvature $$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$$
now $y'=-\frac{b^2}{a^2}{\cdot}\frac{x}{y}$
$y''=\frac{-a^2y'^2-b^2}{a^2y}$
putting up the values in formula
we get
$$\kappa =\frac{\frac{a^2y'^2+b^2}{a^2y}}{\bigg[1+\frac{b^4}{a^4}{\cdot}\frac{x^2}{y^2}\bigg]^\frac{3}{2}} $$
$$\kappa =\frac{\frac{b^4x^2+a^2y^2b^2}{a^2y^3}}{\frac{\bigg[a^4y^2+b^4{\cdot}x^2\bigg]^\frac{3}{2}}{a^2y^3}} $$
now it can be solved
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Derivative of $\tan^{-1}(f(x))$ What is derivative of $$\tan^{-1}\left(\frac{{\sqrt{4+x}+\sqrt{4-x}}}{\sqrt{4+x}-\sqrt{4-x}}\right).$$ So I tried to write it as $\tan(\tan^{-1}(...))$ to get the $f(x)=\frac{\pi}{4}+\tan^{-1}\left(\sqrt{\frac{4+x}{4-x}}\right)$ but still it's not better. Thanks help appreciated
| Easy way :
Put $$x=4\cos 2\theta$$
Then $$y=\tan ^{-1} \left(\frac{\sqrt{4+4 \cos 2\theta}+\sqrt{4-4 \cos 2\theta}}{\sqrt{4+4 \cos 2\theta}-\sqrt{4-4 \cos 2\theta}} \right)$$
$$y=\tan ^{-1} \left(\frac{2\sqrt{1+ \cos 2\theta}+2\sqrt{1- \cos 2\theta}}{2\sqrt{1+ \cos 2\theta}-2\sqrt{1- \cos 2\theta}} \right)$$
$$y=\tan ^{-1} \left(\frac{\sqrt{1+ \cos 2\theta}+\sqrt{1- \cos 2\theta}}{\sqrt{1+ \cos 2\theta}-\sqrt{1- \cos 2\theta}} \right)$$
Since $1+\cos 2\theta =2\cos^2 \theta $ and $1-\cos 2\theta =2\sin^2 \theta$
$$y=\tan ^{-1} \left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta} \right)$$
Now divide both denominator and numerator by $\cos \theta$ ,
$$y=\tan ^{-1} \left(\frac{1+\tan \theta}{1-\tan \theta} \right)$$
$$y=\tan ^{-1} \left(\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4}\cdot\tan \theta} \right)$$
$$y=\tan ^{-1} \left(\tan\left( \frac{\pi}{4}-\theta \right)\right)$$
$$y=\frac{\pi}{4}-\theta$$
So $$\frac{dy}{dx}=0+\frac{d\theta}{dx}$$
Since $x=4 \cos 2\theta$
$$\frac{dx}{d \theta} = -8 \sin 2 \theta $$
So $$\frac{dy}{dx}=\frac{d\theta}{dx}= -\frac{1}{8\sin 2\theta}=-\frac{1}{8 \sqrt{1-\frac{x^2}{16}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Diagonalizability of a given matrix I must find out under which conditions the matrix
$$A=
\left[\begin{array}{ccc|cc}& & & c_0 &\\
& & &c_1&\ddots\\
& & &c_2 &\ddots& c_0\\
& 0 & & \vdots & \ddots &c_1\\
& & & c_d & \ddots &c_2\\
& & & & \ddots & \vdots\\
& & & & & c_d\\
& & & && \end{array}\right]
$$
is diagonalizable (where the $c_d$ lie on the diagonal and the first $d$ columns are $0$, and so everywhere else).
Perhaps the Jordan form is easy to calculate?
| Note that your matrix $A$ is a block upper triangular matrix of the form
$$ A = \begin{pmatrix} 0_{d \times d} & B \\ 0 & D_{(d + 1) \times (d + 1)} \end{pmatrix} $$
where $D$ is also a block upper triangular matrix with $c_d$ on the diagonal. Hence, the characteristic polynomial of $A$ is
$$ p_A(x) = x^d(x - c_d)^{d+1}. $$
We have two options:
*
*If $c_d = 0$ then the characteristic polynomial of $A$ is $x^{2d + 1}$ and so $A$ is nilpotent which implies that if $A$ is also diagonalizable we must have $A = 0$. Hence, $c_d = c_{d-1} = \dots = c_0 = 0$.
*If $c_d \neq 0$ then $A$ will be diagonalizable if and only if the minimal polynomial of $A$ is $x(x - c_d)$. Plugging $A$ into $x(x - c_d)$, we see that we must have
$$ A(A - c_d I) = \begin{pmatrix} 0 & B \\ 0 & D \end{pmatrix}
\begin{pmatrix} -c_d I & B \\ 0 & D - c_d I \end{pmatrix} = \begin{pmatrix} 0 & B(D - c_d I) \\ 0 & D(D - c_d I) \end{pmatrix} = 0. $$
In particular, $D(D - c_dI) = 0$ and so $D$ must be diagonalizable with a single eigenvalue $c_d$ which implies that in fact $D = c_dI$ and so $c_{d-1} = \dots = c_0 = 0$.
In any case, we see that we must have $c_{d-1} = \dots = c_0$ for the matrix to be diagonalizable.
An alternative, more elementary, solution that doesn't involve the minimal polynomial continuous from the calculation of the characteristic polynomial as follows:
*
*If $c_d = 0$ then the characteristic polynomial of $A$ is $x^{2d+1}$ and so if $A$ is diagonalizable, we must have $\dim \ker(A) = 2d + 1$ so $A = 0$.
*If $c_d \neq 0$ then we must have $\dim \ker A = d$ and $\dim \ker (A - c_d I) = d + 1$. Since $c_d \neq 0$, the $d + 1$ non-zero columns of $A$ are linearly independent and so $\dim \ker A = d$. However,
$$ A - c_dI = \begin{pmatrix} -c_dI & B \\ 0 & D - c_dI \end{pmatrix} $$
and so the first $d$ columns of $A - c_dI$ are linearly independent. In order that $\dim \operatorname{Im}(A - c_dI) = d$, the rest of the columns must belong to the span of the first $d$ columns. By looking at the last column, we see the this implies that $c_{d-1} = \dots = c_0 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to factorise $(x-1)^2 - (x-5)^2$? My attempt:
$a = (x-1)$
$c = (x-5)$
$a^2 - c^2$
which is equal to:
$$((x-1) - (x-5))((x-1)+(x-5))$$
But the correct answer is :
$8(x-3)$
Can you explain, please?
| $$\begin{array}{rl} & \begin{bmatrix} -1\\ 1\end{bmatrix} \begin{bmatrix} -1\\ 1\end{bmatrix}^T - \begin{bmatrix} -5\\ 1\end{bmatrix} \begin{bmatrix} -5\\ 1\end{bmatrix}^T = \begin{bmatrix} 1 & -1\\ -1 & 1\end{bmatrix} - \begin{bmatrix} 25 & -5\\ -5 & 1\end{bmatrix}\\\\ &= \begin{bmatrix} -24 & 4\\ 4 & 0\end{bmatrix} = 4 \begin{bmatrix} -6 & 1\\ 1 & 0\end{bmatrix} \equiv 4 \begin{bmatrix} -6 & 2\\ 0 & 0\end{bmatrix} = 8 \begin{bmatrix} -3 & 1\\ 0 & 0\end{bmatrix} = \begin{bmatrix} 8\\ 0\end{bmatrix} \begin{bmatrix} -3\\ 1\end{bmatrix}^T\end{array}$$
Hence,
$$(x-1)^2 - (x-5)^2 = 8 (x-3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$ I am working on the following exercise. Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$.
I have a couple of ideas for going about this exercise.
$1)$ By moving $1$ to the other side of the equation we obtain:
$y^4-1=x^4-x^3+x^2-x \rightarrow (y^2-1)(y^2+1)=x(x-1)(x^2+1)$.
The LHS gives two consecutive integers I am able to see, but other than that I am stuck and not sure where to go from here.
$2)$ We notice that $1-x+x^2-x^3+x^4=\Phi_{10}(x)=y^4$.
We now have that $\Phi_{10}(x) \geq 0$ since $y=\pm \sqrt[4]{1-x+x^2-x^3+x^4}$.
In this case, with some computation, $x={0,1}$ are the only possibilities that will force $y \in \mathbb{Z}$. From there, I use induction to show that $x \geq 2$ and $x \leq -1$ do not yield a perfect fourth.
In either case, I keep running into some issues. If someone could please other a hint to help me continue through this exercise, that would be very helpful. Thank you.
Update:
Hello, my progress on the following problem is as follows:
Suppose that $x=y$, then we get $y^3-y^2+y-1=0 \rightarrow y=1, \pm i$. Thus we get two solutions $(x,y)={(1,1),(1,-1)}$, the $y=-1$ since $y^4$ was in the original equation.
Next, using @Batominovski hint we can write:
$(2x^2-x)^2 < (x^4-x^3+x^2-x+1) < 4(x^4-x^3+x^2-x+1) \leq (2x^2-x+2)^2 \forall x \in \mathbb{R}$ which is true for $x=0$.
Since the multiplication of a constant (in this case $k=4$) does not affect a solution from existing, we have that $x=0$ is a solution, consequently we find two more solutions are $(x,y)=(0,1),(0,-1)$.
| Hint: Note that
$$\left(2x^2-x\right)^2<4\left(x^4-x^3+x^2-x+1\right)\leq\left(2x^2-x+2\right)^2$$
for all $x\in\mathbb{R}$. The right-hand side becomes an equality if and only if $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proving $\lim\limits_{x\rightarrow 1} \frac{x^2+3}{x+1}=2$ using the formal definition of the limit
Prove $\lim\limits_{x\rightarrow 1} \frac{x^2+3}{x+1}=2$ using the formal definition of the limit.
My question is, I've picked $\delta\lt1$, and I've found that $\delta \lt \min(1,\sqrt{\epsilon})$. Was picking $1$ problematic at all? and is my choice for $\delta$ correct?
Rest of Proof:
$$0\lt|x-1|\lt \delta \Rightarrow \left|\frac{(x-1)^2}{x+1}\right|\lt \epsilon$$
Picking $\delta \lt 1$:
$$|x-1|\lt 1 \Rightarrow -1\lt x-1 \lt 1$$
And we get from that $\frac13 \lt \frac{1}{x+1} \lt 1$ which leads to $\left|\frac{1}{x+1}\right| \lt 1$
Let's go back:
$$\left|\frac{(x-1)^2}{x+1}\right|\lt \left|1(x-1)^2\right|\lt \epsilon$$
Since $(x-1)^2\gt 0$ we can get rid of the absolute value
and we get
$$(x-1)^2\lt \epsilon \rightarrow x-1 \lt \sqrt{\epsilon}$$
Also: What is the difference between picking $\delta=1$ and $\delta \lt 1$
| It seems me correct,the alternative short proof could be:
For any $\epsilon >0$ ,and by choosing $\quad \delta =min\left( 1,\epsilon \right) $ and note that if $\left| x-1 \right| <\delta $ the we will get
$$
\left| \frac { x^{ 2 }+3 }{ x+1 } -2 \right| =\left| \frac { { x }^{ 2 }-2x+1 }{ x+1 } \right| =\left| \frac { { \left( x-1 \right) }^{ 2 } }{ \left( x-1 \right) +2 } \right| <\left| \frac { { \left( x-1 \right) }^{ 2 } }{ \left( x-1 \right) } \right| =\left| x-1 \right| <\epsilon
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $x^6+x^4+x^3+x^2+1=0$? There is a hint in the question, use the factorization of $x^5+x+1$.
| Divide by $x^3$ and remember that
$$
x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)
$$
so the equation becomes
$$
\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)
+\left(x+\frac{1}{x}\right)+1=0
$$
Set $t=x+1/x$ and you get
$$
t^3-2t+1=0
$$
Can you go on? Note that $t=1$ is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Prove $ \sin x + \frac{ \sin3x }{3} + ... + \frac{ \sin((2n-1)x) }{2n-1} >0 $ Prove that for $ 0<x< \pi $, $$ \quad S_n(x) = \sin x + \frac{ \sin3x }{3} + ... + \frac{ \sin((2n-1)x) }{2n-1} >0 \quad \forall n = 1,2,... $$
Having trouble with this problem. This is an olympiad-style question, so an answer that doesn't use calculus or analysis would be preferred. A possible approach is induction, but for this we need to find a function in terms of $n$ and $x$ so that we can actually use the inductive step. If anyone has any ideas they would be appreciated.
If you really want to go down the calculus route (at this point I don't mind), then $ S_n' (x) = \cos x + \cos 3x + ... +\cos((2n-1)x) $ , which you can find a closed form for, but I don't know how useful that is.
| If we want to avoid integration, we can obtain the result via summation by parts. Let
$$F_m(x) := \sum_{k = 0}^{m-1} \sin \bigl((2k+1)x\bigr).\tag{1}$$
Using the addition theorem for the cosine, we obtain
$$\sin(rx) = \frac{2\sin x \sin(rx)}{2\sin x} = \frac{\cos\bigl((r-1)x\bigr) - \cos \bigl((r+1)x\bigr)}{2\sin x}$$
and hence by telescoping
$$F_m(x) = \frac{1 - \cos (2mx)}{2\sin x}\tag{2}$$
when $x$ is not an integer multiple of $\pi$. Since $1 - \cos (2mx) \geqslant 0$ for all $x \in \mathbb{R}$ and $m \in \mathbb{N}$, we have $F_m(x) \geqslant 0$ for $0 < x < \pi$ and $m \in \mathbb{N}$. Thus
\begin{align}
S_n(x) &= \sum_{k = 0}^{n-1} \frac{\sin \bigl((2k+1)x\bigr)}{2k+1}\\
&= \sum_{k = 0}^{n-1} \frac{F_{k+1}(x) - F_k(x)}{2k+1}\\
&= \sum_{k = 0}^{n-1} \frac{F_{k+1}(x)}{2k+1} - \sum_{k = 0}^{n-1} \frac{F_k(x)}{2k+1}\\
&= \sum_{m = 1}^n \frac{F_m(x)}{2m-1} - \sum_{k = 1}^{n-1} \frac{F_k(x)}{2k+1}\tag{$F_0 \equiv 0$}\\
&= \frac{F_n(x)}{2n-1} + \sum_{k = 1}^{n-1} \biggl(\frac{1}{2k-1} - \frac{1}{2k+1}\biggr) F_k(x)\\
&\geqslant 0,
\end{align}
since all terms on the right hand side are non-negative. The strict inequality $S_n(x) > 0$ for $x\in (0,\pi)$ and $n \in \mathbb{N}\setminus \{0\}$ follows because the term involving $F_1(x) = \sin x$ is strictly positive on $(0,\pi)$.
If we use integration, we first observe that since $\sin \bigl(m(\pi - x)\bigr) = (-1)^{m+1}\sin x$, we have the symmetry $S_n(\pi-x) = S_n(x)$, so it suffices to consider $0 < x \leqslant \pi/2$. Then we note that
$$S_n'(x) = \frac{\sin (2nx)}{2\sin x},\tag{3}$$
with more or less the same argument as above. Since clearly $S_n(0) = 0$, we have
$$S_n(x) = \int_0^x S_n'(t)\,dt = \int_0^x \frac{\sin (2nt)}{2\sin t}\,dt = \frac{1}{4n}\int_0^{2nx} \frac{\sin u}{\sin \frac{u}{2n}}\,du.$$
The denominator is strictly increasing on the interval of integration (since $0 < x \leqslant \pi/2$), and hence
$$\int_{2k\pi}^{2(k+1)\pi} \frac{\sin u}{\sin \frac{u}{2n}}\,du = \int_{2k\pi}^{(2k+1)\pi} \sin u\biggl( \frac{1}{\sin \frac{u}{2n}} - \frac{1}{\sin \frac{u+\pi}{2n}}\biggr)du > 0\tag{4}$$
for $0 \leqslant k < n$. Let $m = \bigl\lfloor \frac{2nx}{2\pi}\bigr\rfloor$, then
$$4nS_n(x) = \sum_{k = 0}^{m-1} \int_{2k\pi}^{2(k+1)\pi} \frac{\sin u}{\sin \frac{u}{2n}}\,du + \int_{2m\pi}^{2nx} \frac{\sin u}{\sin \frac{u}{2n}}\,du.$$
The last integral is strictly positive - if $2nx \leqslant (2m+1)\pi$, the integrand is strictly positive on the whole interval except the endpoints, if $(2m+1)\pi < 2nx < 2(m+1)\pi$, $(4)$ shows that $\int_{2m\pi}^{2nx}\dotsc\,du > \int_{2m\pi}^{2(m+1)\pi}\dotsc\,du > 0$ - unless $2nx = 2m\pi$. But in that case, we have $m \geqslant 1$, and the sum is strictly positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Finding the coefficient of $x^r$ in an expansion. Suppose that the summation of the infinite series $$1+nx+\frac{n(n-1)}{2} x^2+\cdots+\frac{n(n-1)\cdots(n-r+1)}{r}x^r+\cdots$$ is equal to $(1+x)^n$ for $|x|<1$.
Show that the coefficient of $x^r$ in the expansion of $\frac{1+x+x^2}{(1-x)^2}$ is $3r $.
Hence show that $(217)^\frac{1}{3} \simeq 6.0092$
My attempt :
$$\frac{1+x+x^2}{(1-x)^2}=\frac{(1+x+x^2)(1-x)}{(1-x)^3}$$
$$\frac{1+x+x^2}{(1-x)^2}=\frac{1-x^3}{(1-x)^3}$$
$$=\frac{1}{(1-x)^3}-\frac{x^3}{(1-x)^3}$$
$$=\frac{1}{\left(1+(-x)\right)^3}+\frac{1}{\left(1+\left(-\frac{1}{x}\right)\right)^3}$$
$$=(1+(-x))^{-3}+\left(1+\left(-\frac{1}{x}\right)\right)^{-3}$$
How can I proceed after this ? Is there another method ? Is my method correct ?
| $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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\begin{align}
\color{#f00}{{1 + x + x^{2} \over \pars{1 - x}^{2}}} & =
{1 - x^{3} \over \pars{1 - x}^{3}} =
\pars{1 - x^{3}}\sum_{r = 0}^{\infty}{-3 \choose r}\pars{-x}^{r} =
\half\pars{1 - x^{3}}\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r}
\end{align}
Note that
$\ds{{-3 \choose r} = {-\pars{-3} + r - 1 \choose r}\pars{-1}^{r} =
{r + 2 \choose r}\pars{-1}^{r} =
\half\pars{r + 2}\pars{r + 1}\pars{-1}^{r}}$
\begin{align}
\color{#f00}{{1 + x + x^{2} \over \pars{1 - x}^{2}}} & =
\half\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r} -
\half\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r + 3}
\\[5mm] & =
\half\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r} -
\half\sum_{r = 3}^{\infty}\pars{r - 1}\pars{r - 2}x^{r}
\\[5mm] & =
1 + 3x + 6x^{2} +
\half\sum_{r = 3}^{\infty}\bracks{%
\pars{r + 2}\pars{r + 1} - \pars{r - 1}\pars{r - 2}}x^{r}
\\[5mm] & =
1 + 3x + 6x^{2} + \sum_{r = 3}^{\infty}\pars{3r}x^{r} =
1 + \sum_{r = 1}^{\infty}\pars{3r}x^{r}
\end{align}
$$
\color{#f00}{\bracks{x^{r}}\bracks{{1 + x + x^{2} \over \pars{1 - x}^{2}}}}
=
\color{#f00}{\left\lbrace\begin{array}{rcrcl}
\ds{1} & \mbox{if} & \ds{r} & \ds{=} & \ds{0}
\\[2mm]
\ds{3r} & \mbox{if} & \ds{r} & \ds{=} & \ds{1,2,3,\ldots}
\\[2mm]
\ds{0} &&&& \mbox{otherwise}
\end{array}\right.}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove inequality $\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$
Let $x,y,z>0$ and $x+y+z=1$. Prove that
$$\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$$
My work so far:
I use Titu's Lemma:
$$\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x=$$
$$=\frac{x^4}{xy}+\frac{y^4}{yz}+\frac{z^4}{zx}+\frac{x^2}{\frac zx}+\frac{z^2}{\frac yz}+\frac{y^2}{\frac xy}\ge$$
$$\ge\frac{\left(x^2+y^2+z^2+x+y+z\right)^2}{xy+yz+zx+ \frac zx+\frac yz+\frac xy}=\frac{\left(x^2+y^2+z^2+1\right)^2}{xy+yz+zx+ \frac zx+\frac yz+\frac xy}$$
I need help here.
| Suppose $x \geq y \geq z$. Using the rearrangements inequality we have
$$ \frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x} \geq x^3 \cdot \frac{1}{x}+...+z^3 \cdot \frac{1}{z} =x^2+y^2+z^2 $$
since $x^3,y^3,z^3$ and $1/x,1/y,1/z$ have opposite ordering. The same thing applies to
$$ \frac{x^3}{z}+\frac{z^3}{y}+\frac{y^3}{x} \geq x^2+y^2+z^2$$
Thus the left hand side is greater than $2(x^2+y^2+z^2)$. This is immediately greater than the RHS by noting that $1=(x+y+z)^2$.
You could use "Titu's Lemma" if you like, just like in your try, but split in half:
$$ \frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x} \geq \frac{(x^2+y^2+z^2)^2}{xy+yz+zx} \geq \frac{(x^2+y^2+z^2)^2}{x^2+y^2+z^2} = x^2+y^2+z^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing $\lim_{n \to \infty} \left(\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\cdots+\frac{n}{(n+n)^2}\right)$
Calculate $$\lim_{n \to \infty} \left(\dfrac{n}{(n+1)^2}+\dfrac{n}{(n+2)^2}+\cdots+\dfrac{n}{(n+n)^2}\right).$$
I tried turning this into a Riemann sum, but didn't see how since we get $\dfrac{1}{n} \cdot \dfrac{n^2}{(n+k)^2}$, which I don't see how relates.
| $\frac{n}{n+2}-\frac{n}{2n+1}=n\int\limits_{n+2}^{2n+1}\frac{1}{x^2}dx<n(\frac{1}{n+1}^2+\frac{1}{n+2}^2+\dots + \frac{1}{2n}^2)< n\int\limits_{n+1}^{2n}\frac{1}{x^2}dx=\frac{n}{n+1}-\frac{n}{2n}$
Clearly the expressions on both sides go to $\frac{1}{2}$ when $n\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the remainder of $2^n+n^2$ modulus 6
Find the remainder of $2^n+n^2$ modulus 6 given that $2^n+n^2$ is a prime and $n\geq2$($n$ positive integer)
I tried to solve this but failed!I just know that $n$ must be odd. No progress at all!!
| 2^n is never odd, so n^2 has to be odd to give a prime, n is odd
2^n with n odd is 4^k 2, 4^k = 4 mod 6, so
2^n = 2 mod 6 here
n is 1, 3 or 5 mod 6
n^2 is 1 or 3 mod 6
if n^2 = 1 mod 6, then 2^n + n^2 = 3 mod 6, which would mean that when divided by 6, then 3 would remain, meaning that 2^n + n^2 divides by 3 and is not prime
so n^2 = 3 mod 6
2^n + n^2 = = 5 mod 6
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can you prove Ceva's Theorem using vectors? Given the following layout, where $ \overrightarrow{CY} = \mathbf{a} $ and $ \overrightarrow{CX} = \mathbf{b} $:
Prove that $$ \frac{\overrightarrow{BX}}{\overrightarrow{XC}} \times \frac{\overrightarrow{CY}}{\overrightarrow{YA}} \times \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}} = 1$$
| Our proof is in three parts. Define $ \overrightarrow{CB} = \alpha\overrightarrow{CX}$, $ \overrightarrow{CA} = \beta\overrightarrow{CY}$ and $ \overrightarrow{AB} = \gamma\overrightarrow{AZ}$. Also, let $ \overrightarrow{AX} = \lambda\overrightarrow{AP}$, $ \overrightarrow{BY} = \mu\overrightarrow{BP}$ and $ \overrightarrow{CZ} = \nu\overrightarrow{CP}$.
1. $ \alpha, \beta, \overrightarrow{CP} $
$\overrightarrow{CP} = \overrightarrow{CX} + \overrightarrow{XP} = \mathbf{b} - \overrightarrow{PX} = \mathbf{b} - (\overrightarrow{AX} - \overrightarrow{AP}) = \mathbf{b} - (\overrightarrow{AX} - \frac{1}{\lambda}\overrightarrow{AX}) = \mathbf{b} - (\frac{\lambda - 1}{\lambda}(\overrightarrow{AC} + \overrightarrow{CX})) = $ $\mathbf{b} - \frac{\lambda - 1}{\lambda}\mathbf{b} + \frac{(\lambda - 1)\beta}{\lambda}\mathbf{a}$
By a similar argument, $ \overrightarrow{CP} = \overrightarrow{CY} + \overrightarrow{YP} = \mathbf{a} - \frac{\mu - 1}{\mu}\mathbf{a} + \frac{(\mu - 1)\beta}{\mu}\mathbf{b}$
So, $ \mathbf{b} - \frac{\lambda - 1}{\lambda}\mathbf{b} + \frac{(\lambda - 1)\beta}{\lambda}\mathbf{a} = \mathbf{a} - \frac{\mu - 1}{\mu}\mathbf{a} + \frac{(\mu - 1)\beta}{\mu}\mathbf{b}$
$ (1 - \frac{\lambda - 1}{\lambda} - \frac{(\mu - 1)\alpha}{\mu})\mathbf{b} = (1 - \frac{\mu - 1}{\mu} - \frac{(\lambda - 1)\beta}{\lambda})\mathbf{a}$
From which it follows that $$ 1 - \frac{\lambda - 1}{\lambda} - \frac{(\mu - 1)\alpha}{\mu} = 0 \Rightarrow \beta = \frac{\lambda}{\lambda\mu - \mu} $$
and
$$ 1 - \frac{\mu - 1}{\mu} - \frac{(\lambda - 1)\beta}{\lambda} = 0 \Rightarrow \alpha = \frac{\mu}{\mu\lambda - \lambda} $$
We can substitute either one of these into either expression for $ \overrightarrow{CP}$ to get $$ \overrightarrow{CP} = \frac{1}{\mu}\mathbf{a} + \frac{1}{\lambda}\mathbf{b}$$
2. $ \gamma $
$ \overrightarrow{CZ} = \overrightarrow{CA} + \overrightarrow{AZ} = \beta \mathbf{a} + \frac{1}{\gamma}(\overrightarrow{AC} + \overrightarrow{CB}) = \beta \mathbf{a} - \frac{\beta}{\gamma}\mathbf{a} + \frac{\alpha}{\gamma}\mathbf{b}$
We substitute this result into $ \overrightarrow{CZ} = \nu\overrightarrow{CP} $ to get $$ \beta \mathbf{a} - \frac{\beta}{\gamma}\mathbf{a} + \frac{\alpha}{\gamma}\mathbf{b} = \frac{\nu}{\mu}\mathbf{a} + \frac{\nu}{\lambda}\mathbf{b}$$
$$ (\beta - \frac{\beta}{\gamma} - \frac{\nu}{\mu})\mathbf{a} = (\frac{\nu}{\lambda} - \frac{\alpha}{\gamma})\mathbf{b}$$
From which it follows that $$ \beta - \frac{\beta}{\gamma} - \frac{\nu}{\mu} = 0 \Rightarrow \gamma = \frac{\nu\gamma}{\beta\mu} + 1 $$
and
$$ \frac{\nu}{\lambda} - \frac{\alpha}{\gamma} = 0 \Rightarrow \nu\gamma = \alpha\lambda $$
So, once we substitute our values for $ \alpha $ and $ \beta $ $$ \gamma = \frac{\alpha\lambda}{\beta\mu} + 1 = \frac{\mu(\lambda - 1)}{\lambda(\mu - 1)} + 1 $$
3. Tie it all together
By substituting in our values for $\alpha,\beta $ and $ \gamma $ we achieve
$$
\begin{align}
\frac{\overrightarrow{BX}}{\overrightarrow{XC}} \times \frac{\overrightarrow{CY}}{\overrightarrow{YA}} \times \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}} & = \frac{\alpha - 1}{(\beta - 1)(\gamma - 1)} \\
& = 1
\end{align}
$$
The above proof uses the most fundamental approach of equating two vectors which are the same and solving a pair of simultaneous equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The indefinite integral $ \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$ $$ \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$$
$$ \int \frac{\sin \left(x +\alpha\right)}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}$$
$$ \int \frac{\sin \left(x +\alpha\right)}{\cos^3 x}{\frac{\sin x + \cos x}{\sqrt{\cos 2x}}}$$
$$ \sqrt 2 \int \frac{\sin \left(2x +\alpha+ \frac{\pi}{4}\right) + \sin \left(\alpha -\frac{\pi}{4}\right)}{\cos^3 x\cdot\sqrt{\cos 2x}}$$
How I can do it after this and get rid of square root?
| Let $$I = \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}dx$$
$$I = \cos \alpha\int \frac{\sin x+\cos x\cdot \tan \alpha}{\cos x}\cdot \sqrt{\frac{1+\tan x}{1-\tan x}}\cdot \sec^2 xdx$$
So $$I = \cos \alpha \int (\tan \alpha+\tan x)\sqrt{\frac{1+\tan x}{1-\tan x}}\cdot \sec^2 xdx$$
Now Put $\tan x= t\;,$ Then $\sec^2 xdx = dt$
So $$I = \cos \alpha \int (t+\tan \alpha)\sqrt{\frac{1+t}{1-t}}dt = \cos \alpha \int \frac{(t+\tan \alpha)(t+1)}{\sqrt{1-t^2}}dt$$
So $$I = \cos \alpha \int\frac{t^2+(\tan \alpha +1)t+\tan \alpha}{\sqrt{1-t^2}}dt$$
So $$I = \cos \alpha \int\frac{t^2}{\sqrt{1-t^2}}dt+\cos \alpha \int\frac{(\tan \alpha +1)}{\sqrt{1-t^2}}dt+\cos \alpha\int\frac{\tan \alpha}{\sqrt{1-t^2}}dt$$
In first put $t=\sin \phi\;,$ Then $dt = \cos \phi d\phi$ and in second $(1-t^2)=u^2\;,$ Then $tdt = -udu$
So $$I = \frac{\cos \alpha}{2} \int (1-\cos 2 \phi)d\phi-\cos \alpha (1+\tan \alpha)\int du+\sin \alpha \cdot \sin^{-1}(t)$$
So $$I = \frac{\cos \alpha}{2}\left[\sin^{-1}(t)-t\sqrt{1-t^2}\right]-\cos \alpha(1+\tan \alpha)\sqrt{1-t^2}+\sin\alpha\cdot \sin^{-1}(t)+\mathcal{C}$$
So $$I = \frac{\cos \alpha}{2}\left[\sin^{-1}(\tan x)-t\sqrt{1-\tan^2x}\right]-\cos \alpha(1+\tan \alpha)\sqrt{1-\tan^2x}+\sin \alpha \cdot \sin^{-1}(\tan x)+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1880216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then $|z_{1}+z_{2}+z_{3}|$
If $z_{1},z_{2},z_{3}$ are three complex number such that $|z_{1}| = |z_{2}| = |z_{3}| = 1$
and $\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then possible values of $|z_{1}+z_{2}+z_{3}|$
$\bf{My\; Try::}$ Given $$\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1\Rightarrow z^3_{1}+z_{2}^3+z_{3}^3=-z_{1}z_{2}z_{3}$$
Now Using $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
So $$\left(z_{1}+z_{2}+z_{3}\right)\left[z^2_{1}+z^2_{2}+z^2_{3}-z_{1}z_{2}-z_{2}z_{3}-z_{3}z_{1}\right] = -4z_{1}z_{2}z_{3}$$
So $$(z_{1}+z_{2}+z_{3})\cdot ((z_{1}+z_{2}+z_{3})^2-3(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1})) = -4z_{1}z_{2}z_{3}.......(1)$$
Now Let $k=z_{1}+z_{2}+z_{3}.$ Taking Conjugate, we get $\displaystyle k = \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}} = \frac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}z_{2}z_{3}}$
So we get $z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=kz_{1}z_{2}z_{3}$
Put into $(1)\;,$ We get $$(z_{1}+z_{2}+z_{3})\cdot \left[(z_{1}+z_{2}+z_{3})^2-3kz_{1}z_{2}z_{3}\right] = -4z_{1}z_{2}z_{3}$$
Now How can I solve after that, Help Required, Thanks
| (You made a mistake when taking conjugates: You have $\bar k=\ldots\ $.)
Denote by $s_1$, $s_2$, $s_3$ the elementary symmetric functions of the $z_i$. It is easy to see that we may assume $s_3=1$. Then the $z_i$ are the solutions of the third degree equation
$$z^3-s_1z^2+s_2z-1=0\ .\tag{1}$$
The solutions of the equation $z^3-\bar s_1z^2+\bar s_2z-1=0$ are then the $\bar z_i$, and the solutions of $1-\bar s_1w+\bar s_2 w^2- w^3=0$, resp.
$$w^3-\bar s_2w^2+\bar s_1 w-1=0\tag{2}$$ are the $\bar z_i^{-1}=z_i$ again. Comparing $(1)$ and $(2)$ we see that necessarily $s_2=\bar s_1$.
Now it is well known that the power sum $p_3:=\sum z_i^3$ can be expressed in terms of the $s_j$ as
$$p_3=s_1^3-3s_1s_2+3s_3\ .$$
The given condition then amounts to
$s_1^3-3s_1\bar s_1+3=-1$, or
$$s_1^3=3|s_1|^2-4\ .$$
Writing $s_1=re^{i\phi}$ with $r\geq0$ we see that necessarily $e^{3i\phi}\in\{1, -1\}$, so that either $r^3-3r^2+4=0$, or $r^3+3r^2-4=0$, which then leads to $r\in\{1,2\}$.
Both these values can be realized: Letting $z_1=z_2=-1$, $z_3=1$ the given condition is satisfied with $|z_1+z_2+z_3|=1$, and letting
$z_1=1$, $z_2=e^{i\pi/3}$, $z_3=e^{-i\pi/3}$ leads to $|z_1+z_2+z_3|=2$.
| {
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"url": "https://math.stackexchange.com/questions/1884048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence/Divergence $\int_2^\infty \frac{4x^3+3x^2-x}{5x^5-2x^4+x^2-2}\ln x \, dx$ $$\int_2^\infty \frac{4x^3+3x^2-x}{5x^5-2x^4+x^2-2}\ln x\, dx$$
How should I approach this? can I look at $$\int_2^\infty \frac{\ln x}{x^2}\,dx \text{?}$$
| The integrand function is bounded between $\frac{4\log x}{5 x^2}$ and $\frac{84\log x}{65 x^2}$ on the interval $(2,+\infty)$, hence the integral is converging since
$$ \int_{2}^{+\infty}\frac{\log x}{x^2}\,dx = \frac{1+\log 2}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculate this integral, $\int_0^{\infty}\frac{\ln^2x}{1+x^2}dx$ $\displaystyle\int_0^{\infty}\dfrac{\ln^2x}{1+x^2}dx$
$x=\arctan\alpha$
$dx=\dfrac{1}{1+\alpha^2}d\alpha$
and try a lot items, but didnt arrived anywhere.
| Let us evaluate it via the Mellin transform
\begin{equation}
\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x
\label{eq:1608132}
\tag{2}
\end{equation}
where
\begin{equation}
f(x) = \frac{1}{1+x^{2}}
\label{eq:1608133}
\tag{3}
\end{equation}
Applying the Mellin transform, yields
\begin{equation}
\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} \frac{x^{s-1}}{1+x^{2}} \mathrm{d} x = \frac{\pi}{2}\csc\left(\frac{\pi}{2}s\right)
\label{eq:1608134}
\tag{4}
\end{equation}
And thus
\begin{align}
\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{2}(x)}{1+x^{2}} \mathrm{d} x & = \frac{\mathrm{d}^{2}}{\mathrm{d}s^{2}} \int\limits_{0}^{\infty} \frac{x^{s-1}}{1+x^{2}} \mathrm{d} x |_{s=1} \\
& = \frac{\mathrm{d}^{2}}{\mathrm{d}s^{2}} \frac{\pi}{2}\csc\left(\frac{\pi}{2}s\right) |_{s=1} \\
& = \frac{\pi^{3}}{16}[\cos(\pi s) + 3]\csc^{3}\left(\frac{\pi}{2}s\right) |_{s=1} \\
& = \frac{\pi^{3}}{8}
\label{eq:1608135}
\tag{5}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating the infinite sum of one over the odd numbers squared: $\sum_{i \ge 0} \frac{1}{(2i+1)^2}$ Can someone tell me how to calculate the following infinite sum?
$$
(1/1^2)+(1/3^2)+(1/5^2)+(1/7^2)+(1/9^2)+(1/11^2)+ \cdots
$$
Don't give me the answer. Can you tell me if this is a geometric series?
| To test if it's geometric, compare the ratio of adjacent terms. For example, $\frac{1/3^2}{1/1^2}$ and $\frac{1/5^2}{1/3^2}$. Are these two quantities equal?
Hint:
$$
\left( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \cdots\right) \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \frac{1}{16^2} + \cdots \right) \\
= \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \cdots
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\cos x + \cos y - \cos(x+y) = \frac{3}{2}$, then how are $x$ and $y$ related?
If $$\cos x + \cos y - \cos(x+y) = \frac{3}{2}$$ then
*
*(a) $x + y = 0$
*(b) $x = 2 y$
*(c) $x = y$
*(d) $2 x = y$
It is problem of trigonometry, and I have the solution of the problem. However, after seeing the solution, I don't quite understand how is one supposed to know how to approach this problem. My request is:
Don't just solve the problem, but also tell me from where do you come to know which approach would work. (Tell me from where do you come to know which trick or approach to be used to solution this problem.)
Thanks in advance.
| Let $z_1 = -\cos x + i\sin x$, $z_2 = -\cos y + i\sin y$. Then,
\begin{align*}
|1+z_1+z_2|^2 \geq 0 &\Rightarrow 3 -2\cos x - 2\cos y + 2\cos(x+y) \geq 0\\
&\Rightarrow \cos x + \cos y - \cos(x+y) \leq \frac{3}{2}
\end{align*}
Equality holds if and only if $1 + z_1 + z_2 = 0$ and hence $1+\bar{z_1}+\bar{z_2} =0$. Adding, we obtain, $\cos x + \cos y = 1$ and consequently $\cos(x+y) = -\frac{1}{2}$. Thus $x+y = \frac{2\pi}{3}$ and from $\cos x + \cos y = 1 $, we get $x = \frac{\pi}{3}$ and $y = \frac{\pi}{3}$. Thus $x=y$
This can also be solved without using complex numbers:
\begin{align*}
&\qquad \cos x + \cos y - \cos(x+y) = \frac{3}{2} \\
&\Rightarrow 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) - 2\cos^2\left(\frac{x+y}{2}\right) + 1 = \frac{3}{2} \\
&\Rightarrow 4\cos^2\left(\frac{x+y}{2}\right) - 4\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) +1 = 0 \\
&\Rightarrow \left(2\cos\left(\frac{x+y}{2}\right)-\cos\left(\frac{x-y}{2}\right)\right)^2 + 1 - \cos^2\left(\frac{x-y}{2}\right) = 0\\
&\Rightarrow \left(2\cos\left(\frac{x+y}{2}\right)-\cos\left(\frac{x-y}{2}\right)\right)^2 + \sin^2\left(\frac{x-y}{2}\right) = 0
\end{align*}
and hence
\begin{align*}
\sin\left(\frac{x-y}{2}\right) &= 0\\
2\cos\left(\frac{x+y}{2}\right)-\cos\left(\frac{x-y}{2}\right) &= 0
\end{align*}
Thus $x=y = \dfrac{\pi}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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solution of absolute value equation Question: If $x + |x| + y = 7$ and $x + |y| - y = 6$, then $x + y =?$
My solution:
I considered each cases of $x$ and $y$ i.e $x$ positive and $y$ positive, $x$ positive and $y$ negative, $x$ negative and $y$ positive, and finally $x$ negative and $y$ negative. After solving I found $x= 4$ and $y=-1$ is the solution. So, their sum is $4-1= 3$.
Is there any smarter way to solve this type of equation?
| Substitute $y=7-x-|x|$, giving an equation in a single unknown, $$x+|7-x-|x||-(7-x-|x|)=6.$$
Then for $x\le0$, $x=6$, which is impossible and for $x\ge0$, $x+|7-2x|-(7-2x)=6$.
You still have to distinguish $2x\le7$, giving $x=6$, which is impossible, and for $2x\ge7$, $x-14+4x=6$ or $$\color{green}{x=4,y=-1}.$$
Not much simpler.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$ Could it be possible to find the solution for the following series?
$$ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Thanks in advance!
| Let $$S = 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Also we know, $$(1-b)(1+b+b^2+b^3 \cdots +b^n-1) = 1-b^n$$
We now start feeling that having $(1-b)$ as a factor in each term in the $RHS$ will create a "nice" series.
So multiplying the $LHS$ & $RHS$ by $(1-b)$, we get, $$S(1-b) = (1-b)+a(1-b^2)+a^2(1-b^3)+a^3(1-b^4)+\cdots $$
Now, upon expanding the $RHS$, we get 2 geometric progressions,
$$S(1-b) = (1+a+a^2+a^3+\cdots)-(b+ab^2+a^2b^3+a^3b^4+\cdots)$$
The first series is an infinite GP ( here $|a|<1$ is a constraint) with the first term equal to $1$ and common ratio $a$. So its sum, $S_1$ (say) is given by,
$$S_1 = \frac{1}{1-a}$$
The second series is also an infinite GP ( here too $|ab|<1$ is a constraint). It's first term is $b$ and the common ratio is $ab$. It's sum, $S_2$ (say), is given by,
$$S_2=\frac{b}{1-ab}$$
Combining $S_1$ and $S_2$ we finally get,
$$S(1-b) = \frac{1}{1-a} - \frac{b}{1-ab}$$
Simplifying the $RHS$ and dividing both sides of the equation by $(1-b)$ we get,
$$S = \frac{1}{(1-a)(1-ab)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Box contains 2 different coins, one is chosen randomly and tossed n times, head came all n times. Probability that n+1 toss is a head too? A box contains two coins: a regular coin and one fake two-headed coin (P(H)=1). One coin is choose at random and tossed $n$ times.
If the first n coin tosses result in heads, What is the probability that the $(n+1)^{th}$ coin toss will also result in heads?
My solution:
$$\text{Required Probability} = \frac{1}{2}\times \left(\frac{1}{2}\right)^n \times \frac{1}{2}+\frac{1}{2} \times 1^n \times 1$$
| The probability of seeing $n$ heads in $n$ tosses of a fair coin is $2^{-n}$. Thus the total probability of seeing $n$ consecutive heads is $$\frac 12\times 2^{-n}+\frac 12 \times 1$$ Therefore, given that you have in fact observed $n$ consecutive heads the new estimate for the probability that you have the fair coin is $$\frac {\frac 12 \times 2^{-n}}{\frac 12\times 2^{-n}+\frac 12 \times 1}=\frac {1}{1+2^n}$$ and the probability that you have the fake coin is therefore $$\frac {2^n}{1+2^n}$$ It follows that the probability that the next toss is also heads is $$\boxed {\frac 12\times \frac {1}{1+2^n}+\frac {2^n}{1+2^n}=\frac {1+2^{n+1}}{2+2^{n+1}}}$$
Note: as $n$ goes to $\infty$ this tends to $1$, as it clearly should.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$2^n > n^4$ proof by induction This is what I came up with so far:
Inductive step: assume $2^n > n^4$.
Need to prove $2^{n+1} > (n+1)^4$
$$
2^{n+1} = 2 \cdot 2^n > 2 \cdot n^4\\
(2 \cdot n^4)^{1/4} = (2)^{1/4} \cdot n > n+1 \implies 2n^4 > (n+1)^4 \implies 2^n > (n+1)^4
$$
Is there a better way to solve this problem?
| Using the function $x^{1/4}$ is not a purely algebraic proof. Here is one. Explicitely, we'll prove $2^n>n^4$ for all $n>16$.
For that, we'll prove by induction that if $n\ge 16$ and $2^n\ge n^4$, then $2^{n+1}>(n+1)^4$.
For $n=16$, we have an equality: $2^{16}=16^4$.
Now suppose that, for some $n\ge 16$, we have $2^n>n^4$. Then $2^{n+1}=2\cdot 2^n\ge 2n^4$. So we have to prove $2n^4>(n+1)^4$, i.e.
$$n^4>4n^3+6n^2+4n+1.$$
As $n>1$, $\;4n^3+6n^2+4n+1>4n^3+6n^3+4n^3+n^3=15n^3$.
So it is enough to prove $n^4>15n^3$, which is equivalent to $n>15$. We've supposed $n\ge 16$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral ${\large\int}_0^1\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}$ Mathematica claims that
$${\large\int}_0^1\!\!\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}=\frac{\sqrt\pi}{2^{\sqrt2}\sqrt2}\cdot\frac{\Gamma\left(\frac1{\sqrt2}\right)}{\Gamma\left(\frac12+\frac1{\sqrt2}\right)},\tag{$\diamond$}$$
and it also confirms numerically.
How can we prove $(\diamond)$?
This result seems interesting, because no such nice answer seems to occur for other algebraic powers, except trivial cases, when the antiderivative is an elementary function, e.g.
$${\large\int}\frac{dx}{(1+x^\alpha)^\alpha}=\\
\small\frac{x\left(x^\alpha+1\right)^{1-\alpha}}{18}\left[\vphantom{\large|}\left(15(\alpha-1)+4\left(5(\alpha+1)+\left(6\alpha+(\alpha-3)x^{2\alpha}+2(\alpha+2)x^\alpha-3\right)x^\alpha\right)x^\alpha\right)x^\alpha+18\right],$$
where $\alpha=3+\sqrt{10}$.
| Using Euler's integral representation of the Gaussian hypergeometric function, and assuming $a>0$, we get
$$ \begin{align} \int_{0}^{1} \frac{dx}{(1+x^{a})^{a}} &= \frac{1}{a} \int_{0}^{1} \frac{u^{1/a-1}}{(1+u)^{a}} \, du \\ &= \frac{1}{a} \int_{0}^{1} u^{1/a-1} (1-u)^{(1+1/a)-1/a-1}(1+u)^{-a} \, du \\ &= \frac{1}{a} \, B \left(\frac{1}{a}, 1 \right) {}_2F_{1} \left(a, \frac{1}{a}; 1+ \frac{1}{a}; -1 \right) \\&= {}_2F_{1} \left(a, \frac{1}{a}; 1+ \frac{1}{a}; -1 \right). \end{align}$$
In the case $a=\sqrt{2}$, we can apply Kummer's theorem since $1+\sqrt{2} - \frac{1}{\sqrt{2}}= 1+ \frac{1}{\sqrt{2}}$.
$$\begin{align} \int_{0}^{1} \frac{dx}{(1+x^{\sqrt{2}})^{\sqrt{2}}} &= \frac{\Gamma\left(1+ \frac{1}{\sqrt{2}}\right) \Gamma\left(1+\frac{1}{\sqrt{2}}\right)}{\Gamma\left(1+ \sqrt{2}\right) \Gamma\left(1\right)} \\ &= \frac{\Gamma \left(1+ \frac{1}{\sqrt{2}} \right)^{2}}{\Gamma(1+\sqrt{2})}\\ &= \frac{1}{2 \sqrt{2}} \frac{\Gamma\left(\frac{1}{\sqrt{2}}\right)^{2}}{\Gamma(\sqrt{2})} \\ &\approx 0.6604707226 \end{align}$$
To get it in the form given by Mathematica, apply the duplication formula to $\Gamma \left(\frac{1}{\sqrt{2}} \right)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof for $\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$ I am stuck at the following exercise:
Prove that
$\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$.
I tried to prove the Expression by induction but I cannot find a way to prove the implication
$$7\mid(1 + 2^{2^n} + 2^{2^{n+1}}) \Rightarrow 7\mid(1 + 2^{2^{n+1}} + 2^{2^{(n+1)+1}}).$$
Any help would be appreciated very much.
| Suppose $1+2^{2^n}+2^{2^{n+1}}=7k$ and set, for simplicity, $a=2^{2^n}$. Then $2^{2^{n+1}}=a^2$ and $2^{2^{n+2}}=a^4$. Then
\begin{align}
&1+2^{2^n}+2^{2^{n+1}}=1+a+a^2 \\[4px]
&1+2^{2^{n+1}}+2^{2^{n+2}}=1+a^2+a^4
\end{align}
Then
$$
(1+a^2+a^4)-(1+a+a^2)=a^4-a=a(a-1)(a^2+a+1)=7ka(a-1)
$$
so
$$
1+a^2+a^4=7k+7ka(a-1)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What's the best bound for $\sum_{i=0}^{[\lg{n}]}{\binom{n}{2^{i}}}$ I want to find best lower bound and upper bound for:
$$\sum_{i=0}^{[\lg{n}]}{\binom{n}{2^{i}}}$$
| As @JackD'Aurizio states, the sum is dominated by the choices of $i$ for which $2^i$ is closest to $\frac{n}{2}$. Let $a = \left\lfloor \lg \frac{n}{2} \right\rfloor = \lfloor \lg n \rfloor - 1$. This gives the largest $i$ for which $2^i \le \frac{n}{2}$. The terms we should focus on are then $i = a$ and $i = a + 1$.
(To see that the other terms are insignificant, observe $\binom{n}{r-1} = \frac{r}{n-r} \binom{n}{r} \le \frac12 \binom{n}{r}$, provided $r \le \frac{n}{3}$. Moreover, since $\frac{n}{4} < 2^a \le \frac{n}{2}$, we have at least $\frac{n}{12}$ values of $r$ for which $2^{a-1} < r \le \frac{n}{3}$. Thus for each of the summands with $i < a$, we have
$$ \binom{n}{2^i} \le \binom{n}{2^{a-1}} = \binom{n}{2^a} \prod_{r = 2^{a-1} + 1}^{2^a} \frac{\binom{n}{r-1}}{\binom{n}{r}} \le 2^{- n/12 } \binom{n}{2^a}. $$
Hence the sum $\sum_{i < a} \binom{n}{2^i} < \lg n \cdot 2^{- n / 12} \binom{n}{2^a}$ is exponentially small compared to $\binom{n}{2^a}$.)
Now note that we have $a = \lfloor \lg n \rfloor - 1 = \lg n - \{ \lg n \} - 1$, where for $x \in \mathbb{R}$, $\{ x\}$ is the fractional part of $x$. Hence $2^a = 2^{\lg n - \{ \lg n \} - 1} = cn$, where $c = 2^{ - \{ \lg n \} - 1 }$. We also have $2^{a+1} = 2cn$.
Now, using Stirling's Approximation, when $k, n \rightarrow \infty$, one has
$$ \binom{n}{k} \sim \sqrt{ \frac{n}{2 \pi k (n-k) } } \left( \frac{n}{k} \right)^k \left( \frac{n }{n-k} \right)^{n-k}. $$
When $k = \gamma n$, for some $\gamma \in (0,1)$, this simplifies to
$$ \binom{n}{\gamma n} \sim \left( 2 \pi \gamma (1 - \gamma) n \right)^{- 1/2} \left( \gamma^{\gamma} (1 - \gamma)^{1 - \gamma} \right)^{-n}.$$
Putting everything together, it follows that
$$ \sum_{i=0}^{\lfloor \lg n \rfloor} \binom{n}{2^i} \sim \left( 2 \pi c (1 - c) n \right)^{- 1/2} \left( c^c (1 - c)^{1 - c} \right)^{-n} + \left( 4 \pi c (1 - 2c) n \right)^{- 1/2} \left( (2c)^{2c} (1 - 2c)^{1 - 2c} \right)^{-n},$$
where $c = 2^{ - \{ \lg n \} - 1}$.
Now of course that is a horrible expression. What is more useful is to compare this to $2^n$. The binary entropy $H(x)$ is defined by $H(x) = -x \lg x - (1-x) \lg (1 - x)$. Then the term $ \left( c^c (1-c)^{1-c} \right)^{-n} = 2^{H(c) n}$, and similarly the other term is $2^{H(2c)n}$. The leading square root factors are insignificant compared to these exponential terms. Hence we find that the order of magnitude of the sum is given by
$$ \sum_{i=0}^{\lfloor \lg n \rfloor} \binom{n}{2^i} = \Theta \left( n^{-1/2} 2^{ f(n) n } \right), $$
where $f(n) = \max \left( H(c), H(2c) \right)$, where $c = 2^{- \{ \lg n \} - 1}$.
Below is a graph showing the values $f(n)$ takes as $\{ \lg n \}$ ranges from $0$ to $1$ (thanks, WolframAlpha!):
So you can see when $\{ \lg n \} \approx 0,1$ (that is, when $n$ is close to a power of $2$), we can get $2^i \approx \frac{n}{2}$, which means the order of magnitude is roughly $2^n / \sqrt{n}$. This is not too far from the trivial upper bound of $2^n$.
On the other hand, for more moderate values of $\{ \lg n \}$ (that is, when $n$ is far from a power of $2$), the constant in the exponent drops. The minimum constant is achieved when $\{ \lg n \} \approx 0.585$, when $2^a \approx \frac{n}{3}$ and $2^{a+1} \approx \frac{2n}{3}$ are both simultaneously as far from $\frac{n}{2}$ as possible. Here the sum is of the order of magnitude roughly $2^{0.918n} / \sqrt{n}$, which is of course exponentially smaller than $2^n$.
[Sorry I can't give a more elegant answer, but these binomial calculations usually require getting your hands dirty.]
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple determinant calculation
\begin{vmatrix}
1 & -2 & 2 \\
1 & -2 & -2\\
2 & -2 & 1
\end{vmatrix}
It is fairly easy I just want to reassure the steps are correct:
\begin{vmatrix}
1 & -2 & 2 \\
1 & -2 & -2\\
2 & -2 & 1
\end{vmatrix} Taking out $-2^{n}$ when $n$ is the number of rows/columns in this case $n=1$ (the second column)
$$-2 \begin{vmatrix}
1 & 1 & 2 \\
1 & 1 & -2\\
2 & 1 & 1
\end{vmatrix} $$
Reducing column 2 from column 1 as no effect on the determinant
$$-2 \begin{vmatrix}
0 & 1 & 2 \\
0 & 1 & -2\\
1 & 1 & 1
\end{vmatrix} $$
Using Laplace expansion on the first column
$$(-2)*1[1*(-2)-2*1]=8.$$
| \begin{vmatrix}
1 & -2 & 2 \\
1 & -2 & -2\\
2 & -2 & 1
\end{vmatrix}
$$R'_1=R_1-R_2 \space R'_2=R_2-R_3 $$
$$\begin{vmatrix}
0 & 0 & 4 \\
-1 & 0 & -3\\
2 & -2 & 1
\end{vmatrix}
=8$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find ordered positive pairs of integers (x, y) that $x^2y^2-2(x+y)$ is a perfect square. Source: Q1 level Hard test 3 of Violympic National Round 2015-16 (Test 3/c/Q1)
(ended in May 2016).
I have looked for similar questions here on Math.SE, but NOTHING was close to mine.
The competition actually asked for the number of solutions, and the answer was 2, but searching here on Wolfram|Alpha shows that there are two solutions $$(x, y) = \{(1, 3), (3, 1)\}$$
Do I have to solve $x^2y^2-2(x+y) = z^2$ (z is an integer) for x or change variables or check possible cases? If it was 1-variable equation like here or here it would simpler.
Any help will be appreciated. Thank you for stopping by. Have a nice day!
| Let us write $x^2y^2 -2(x+y) = (xy-a)^2$. Then, $a(2xy-a) = 2(x+y)$. This means that $a$ must be even, because $2(x+y)$ is even, and $a$ and $2xy-a$ have the same parity.
To give you an intuition about our idea, let us think about the following question: What is the difference of squares of consecutive numbers, or consecutive even numbers, or numbers of an arithmetic progression, as they get larger? The answer is quite simple: As the numbers get larger, the differences get larger.
Now, suppose you have a very large number, and you are to subtract something from it to get a square, as is happening here (with $x^2y^2$ being a very large number, $2(x+y)$ being the subtrahend, and $(xy-a)^2$ being the resulting square). Whatever is being subtracted needs to be large enough to be able to get to the square nearest to the large number, isn't it? This idea will be key to our proof.
Now, think about this: from a very large number, $x^2y^2$, we are subtracting a very small quantity, $2(x+y)$. Will this quantity be enough to reach the previous square? Note that above I just proved that $a$ has to be even, hence the largest square smaller than $x^2y^2$ which we can possibly reach by subtracting $2(x+y)$ is $(xy-2)^2$. All other squares which can possibly be reached by subtracting $2(x+y)$ all have to be smaller than $(xy-2)^2$. Hence, it is wise to check the case $a=2$. That is, we are asking the question: For what $x$ and $y$ is $2(x+y)$ so large enough that subtracting it from $x^2y^2$ can lead to $(xy-2)^2$?
For $a=2$, the difference we are looking to overcome is $a(2xy-a) = 2(2xy-2) = 4xy-4$. We are hence supposed to find all $x$ and $y$ such that $2(x+y)$ is larger than $4xy-4$, for then it will be large enough to result in a perfect square when subtracted, whether that perfect square be $(xy-2)^2$ or $(xy-4)^2$ or $(xy-72)^2$. We are just looking for whether it is large enough to cross the threshold or not.
Now, a chain of equalities: $2(x+y) \geq 4xy-4 \iff 2y+4 \geq x(4y-2) \iff x \leq \frac{2y+4}{4y-2} =\frac{y+2}{2y-1} =\frac{1}{2}+\frac{5}{2(2y-1)}$.
Now, we can verify that $x \leq 3$, because the maximum of the right hand side is $3$, achieved when $y=1$.
Thus, forget about large values of $a$, for $a=2$ itself, only small values $(x \leq 3)$ are guaranteed to give you a large enough value of $2(x+y)$ such that upon subtraction, it is able to reach the previous square. Which is why we do not need to go for general value of $a$ : if $2(x+y)$ was large enough to reach $(xy-a)^2$, then surely it was also large enough to reach $(xy-2)^2$, wasn't it? But then, that again could happen only when $x=3$! That's why I don't bother about general $a$, and start examining values of $x$!
Now we can individually verify the solutions:
Take $x=1$. Then $x^2y^2-2(x+y) = y^2-2y-2 = (y-1)^2 - 3$. This a square only when the difference between two squares is three, which happens only with $1^2$ and $2^2$. Hence, $y-1=2$ and $y=3$.
Take $x=2$. Then $x^2y^2-2(x+y) = 4y^2-2y-4$. Now, I claim that $4y^2-2y-4$ can never be the square of a positive number. For if $4y^2-2y-4=z^2$, then:
$$
y=\frac{1+\sqrt{(2z)^2+17}}{4}
$$
The only two squares differing by $17$ are $9^2=81$ and $8^2=64$. But then $y=\frac{10}{4}$ in this case, hence we do not get an integer value of $y$. Hence there is no solution for $x=2$.
Take $x=3$.Then $x^2y^2-2(x+y) = 9y^2-2y-6 $. Now, I claim that $9y^2-2y-6$ can never be the square of a positive number, except for $y=1$. For if $9y^2-2y-6=z^2$, then:
$$
y=\frac{1+\sqrt{(3z)^2+55}}{9}
$$
All pairs of squares differing by $55$ are the following: $8^2-3^2=55,28^2-27^2=55$. The first case gives $y=\frac{1+8}{9} =1$ and the second case gives $y = \frac{1+28}{9} = \frac{29}{9}$, not an integer. Hence $y=1$ is the only value.
Hence we conclude that $x=1,y=3$ and $x=3,y=1$ are the only permissible values.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{2a^2-1}{b^2+2}$ is not an integer
Let $a$ and $b$ be integers. Prove that $\frac{2a^2-1}{b^2+2}$ is not an integer.
I determined that since $2a^2-1 \equiv 1,7 \pmod{8}$ we must have $b^2+2 \equiv 3 \pmod{8}$ in order for the fraction to be an integer. I didn't see how to find a contradiction from here.
| $b^2+2 \equiv 2,3,6 \pmod{8}$ because $b^2+2$ satisfying $b^2+2 \equiv 3 \pmod{8}$ gets an inverse under module condition. Otherwise, $b^2+2$ will be an even which contradict to the parity of $2a^2-1$
Notice that $3 ^{-1} \equiv 3 \pmod 8$.
And $3 \times 1 \equiv 3^{-1} \pmod 8$, $7 \times 3^{-1} \equiv 5 \pmod 8$ none of which equivalent to $0$.
Here is the contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this integral $\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$ Turns out this integral has a very nice closed form:
$$\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$$
I found it with Mathematica, but I can't figure out how to prove it.
The integral seems quite problematic to me. If the limits were finite, I would do this:
$$\frac{1}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{1}{a^4-b^4}(\sqrt{x^4+a^4}-\sqrt{x^4+b^4})$$
Then, for one of the integrals we will have:
$$\int_A^B \sqrt{x^4+a^4} dx=a^3 \int_{A/a}^{B/a} \sqrt{1+t^4} dt$$
This integral is complicated, but quite well known.
On the other hand $\int_0^\infty \sqrt{1+t^4}dt$ diverges, so I can't consider the two terms separately.
But the integral behaves like I can! If we look at the final expression, it seems like $\int_0^\infty \sqrt{1+t^4}dt=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}}$ even though it can't be correct.
I have to somehow arrive at Beta function, since we have a squared Gamma as an answer.
I'm interested in this integral, since it represents another kind of mean for two numbers $a$ and $b$. If we scale it appropriately:
$$I(a,b)=\frac{8 \sqrt{\pi}}{\Gamma(1/4)^2 } \int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}= \frac{4}{3} \frac{a^2+ab+b^2}{a^3+ab(a+b)+b^3}$$
So, $1/I(a,b)$ is a mean for the two numbers.
| Another way to do it, may be.
Assuming $a>0$ and $t>0$, we have
$$I_a=\int_0^t \sqrt{x^4+a^4}\, dx=a^2 t \, _2F_1\left(-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{t^4}{a^4}\right)$$ Expanding as Taylor series for infinitely large values of $t$, $$I_a=\frac{t^3}{3}+\frac{\sqrt{\frac{\pi }{2}} a^3 \Gamma
\left(\frac{5}{4}\right)}{\Gamma \left(\frac{7}{4}\right)}-\frac{a^4}{2 t}+O\left(\frac{1}{t^3}\right)$$ So, for an infinite value of $t$,
$$\lim_{t\rightarrow\infty}(I_a-I_b)=(a^3-b^3)\frac{\sqrt{\frac{\pi }{2}} \Gamma
\left(\frac{5}{4}\right)}{\Gamma \left(\frac{7}{4}\right)}=(a^3-b^3)\frac{\Gamma \left(\frac{1}{4}\right)^2}{6 \sqrt{\pi }}$$ and hence the result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this a new method for finding powers? Playing with a pencil and paper notebook I noticed the following:
$x=1$
$x^3=1$
$x=2$
$x^3=8$
$x=3$
$x^3=27$
$x=4$
$x^3=64$
$64-27 = 37$
$27-8 = 19$
$8-1 = 7$
$19-7=12$
$37-19=18$
$18-12=6$
I noticed a pattern for first 1..10 (in the above example I just compute the first 3 exponents) exponent values, where the difference is always 6 for increasing exponentials. So to compute $x^3$ for $x=5$, instead of $5\times 5\times 5$, use $(18+6)+37+64 = 125$.
I doubt I've discovered something new, but is there a name for calculating exponents in this way? Is there a proof that it works for all numbers?
There is a similar less complicated pattern for computing $x^2$ values.
| What you have discovered is a finite difference calculation. For any function $f$, in this case the third-power function
$$
f(n) = n^3
$$
we can define the forward difference, or forward discrete derivative:
$$
\Delta f(n) = f(n+1) - f(n) = 3n^2 + 3n + 1
$$
Likewise,
\begin{align*}
\Delta \Delta f(n) = \Delta^2 f(n) &= 6n+ 6 \\
\Delta^3 f(n) &= 6 \\
\Delta^4 f(n) &= 0.
\end{align*}
Your computation,
$$
5^3 = 64 + 37 + 18 + 6
$$
is the statement
$$
f(5) = f(4) + \Delta f(3) + \Delta^2 f(2) + \Delta^3 f(1),
$$
or more generally
$$
f(n) = f(n-1) + \Delta f(n-2) + \Delta^2 f(n-3) + \Delta^3 f(n-4).
$$
This is one discrete analogue of Taylor series
(the more common analogue is Newton's series).
The reason it works is that, for $f(n) = n^3$, $\Delta^4$ and beyond are all zero. So the summation stops once we get to $\Delta^3$.
EDIT:
Let me add that there is yet another identity here which resembles Taylor series, namely,
$$
f(n-1) = f(n) - \Delta f(n) + \Delta^2 f(n) - \Delta^3 f(n).
$$
Altogether, at least when $f$ is a polynomial, we thus have the following Taylor series-like formulas:
\begin{align*}
f(n-1) &= \sum_{i = 0}^\infty (-1)^i [\Delta^i f](n) \\
f(n+1) &= \sum_{i = 0}^\infty [\Delta^i f](n - i) \\
f(n+x) &= \sum_{i=0}^\infty \binom{x}{i} [\Delta^i f](n).
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality solving How must I proceed to solve inequalities of the following form:
$$-1≤r-1/r≤1$$
This is the final step for another problem that I'm attempting to solve and thus far, this is where I've gotten:
By taking the L.C.M and multiplying both sides by $r$ (given to take strictly positive (or zero) values), I got-
$$-r≤r²-2≤2r$$
Taking the left inequality separately, I found that $$(r-1)(r+2) \geq 0$$
which gives me
$$r \in (-\infty, -2] \cup [1, \infty)$$
(Clearly, the only valid region is $[1,\infty)$, since $r$ can only take strictly positive/zero values but let me leave that there until the very end, making a mental note of it.)
The right half of the inequality gives me $r²-2r-2≤0$ and I have no insights as to how to solve this.
Please help! Thanks in advance :) Regards.
| Let's consider two cases, $r > 0$ and $r < 0$.
Case 1
Suppose $r > 0$. Multiplying both sides of the inequality by $r$,
$$-r \leq r^2-1 \leq r.$$
Rearranging the left-side inequality,
$$r^2+r-1 \geq 0.$$
Using the quadratic equation to find the roots of $r^2+r-1$,
$$r = \frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2} .$$
Thus, our inequality becomes,
$$\left(r-\frac{-1 + \sqrt{5}}{2}\right)\left(r-\frac{-1 - \sqrt{5}}{2}\right) \geq 0.$$
From here, we get $r \in (-\infty, \frac{-1 - \sqrt{5}}{2}]$ or $r \in [\frac{-1 + \sqrt{5}}{2}, \infty).$ However, we assumed $r > 0$, thus we only have $r \in [\frac{-1 + \sqrt{5}}{2}, \infty).$
Now let's rearrange the right hand side of the inequality to get
$$r^2-r-1 \leq 0.$$
The roots of $r^2-r-1$ are,
$$\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}.$$
Thus, the right hand side inequality becomes,
$$\left(r-\frac{1 + \sqrt{5}}{2}\right)\left(r-\frac{1 - \sqrt{5}}{2}\right) \leq 0.$$
From here, we get $r \in [\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}]$. But remember $r > 0$, so this means $r \in (0, \frac{1 + \sqrt{5}}{2}]$.
Case 2
Let's multiply both sides of the original inequality by $r$, but remember to flip the direction of the inequalities,
$$-r \geq r^2 - 1 \geq r.$$
Rearranging the left and right side inequalities,
$$r^2+r-1 \leq 0 \text{ and } r^2-r-1 \geq 0.$$
These inequalties are the same as before except with the direction of the inequalties switched around so I'm not going to go through the same level as detail as in case 1.
For the left-hand inequality, we end up with
$r \in [\frac{-1 - \sqrt{5}}{2}, \frac{-1 + \sqrt{5}}{2}]$; however, since $r < 0$, this becomes $r \in [\frac{-1 - \sqrt{5}}{2}, 0)$. For the right-hand inequality, we end up with $r \in (-\infty, \frac{1 - \sqrt{5}}{2}]$ or $r \in [\frac{1 + \sqrt{5}}{2}, \infty).$ However; since $r < 0$, this becomes just $r \in (-\infty, \frac{1 - \sqrt{5}}{2}].$
Putting everything together
Looking at the result of everything, we have,
$$\left(r \in [\frac{-1 + \sqrt{5}}{2}, \infty) \text{ AND } r \in [0, \frac{1 + \sqrt{5}}{2}] \right) \text{ OR } \left( r \in [\frac{-1 - \sqrt{5}}{2}, 0) \text{ AND } r \in (-\infty, \frac{1 - \sqrt{5}}{2}] \right).$$
More succintly,
$$r \in \left([\frac{-1 + \sqrt{5}}{2}, \infty) \cap [0, \frac{1 + \sqrt{5}}{2}] \right) \cup \left([\frac{-1 - \sqrt{5}}{2}, 0) \cap (-\infty, \frac{1 - \sqrt{5}}{2}] \right)$$
$$= [\frac{-1 + \sqrt{5}}{2},\frac{1 + \sqrt{5}}{2}] \cup [\frac{-1 - \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}]$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $x+y=xy=3$ how would you evaluate $x^4+y^4$? I know how to evaluate $x^3 +y^3$ when $x+y=xy=3$, but how would you evaluate for?
$$x^4+y^4$$
Any help would be appreciated.Thanks in advance!
| We have $$81=3^4=(x+y)^4\\=x^4+4x^3y+6x^2y^2+4xy^3+y^4\\=x^4+12x^2+54+12y^2+y^4\\=x^4+y^4+12(x^2+y^2)$$so once you solve for $x^2+y^2$, it should be straight-forward to solve for $x^4+y^4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Differentiating this inverse trigonometric function $$\sin^{-1}\left( \frac{2^{x+1}\cdot3^x}{1+36^{x}} \right)$$
Had this question for todays test but still cannot find out how to proceed.
| If we have $y = \arcsin(2^{x + 1} 3^x / (1 + 36^x))$, then $$\sin y = \frac{2^{x + 1} 3^x}{1 + 36^x}$$ The expression on the right is still a daunting product/quotient, but it can be reduced to sums/differences using logarithms, giving $$\ln \sin y = (x + 1) \ln 2 + x \ln 3 - \ln(1 + 36^x)$$ Now it is routine to take the derivative with respect to $x$ (remember the chain rule for the left!): $$ \frac{y' \cos y}{\sin y} = \ln 2 + \ln 3 - \frac{36^x \ln(36)}{1 + 36^x} = \ln 6 \left(1 - \frac{2 \cdot 36^x}{1 + 36^x}\right) $$ and solving for $y'$ gives $$ y' = \ln 6 \tan y \left(\frac{1 - 36^x}{1 + 36^x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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