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Finding an explicit formula for $a_n$ defined recursively by The sequence $a_n$ is defined recursively by $a_0=0$, $a_n=4a_{n-1}+1$. I must use generating functions to solve this. $n\geq1$.
I have found a pattern:
$$\sum_{n=1}^\infty(4a_{n-1}+1)x^n = x+5x^2+21x^3+85x^4+341x^5+\ldots$$
If we subtract 1 from each term, respectively (ignoring the first term) we would have:
$$x+4x^2+20x^3+84x^4+340x^5+\ldots$$
As you can see, the pattern is the second term increases by 4, the third term increases by 4^2, the fourth term increases by 4^3, and the fifth term increases by 4^4. I wanted to use the formula for the geometric series, but the common ratio isn't 4, but $4^n$. Is there any way I can use this pattern to help me find the explicit formula, using generating functions?
Using $$\sum_{n=0}^\infty 4^nx^n = \frac{1}{1-4x}$$ Isn't helping me too much. But I have a feeling I can manipulate this generating function to solve this.
However, this might seem helpful:
$$\sum_{n=0}^\infty4^nx^{n+1} = 4^0x+4^1x^2+4^2x^3+4^3x^4+\ldots = x+4x^2+16x^3+64x^4+\ldots$$
After some work I have gotten this:
$$A(x)=\frac{x^2}{(1-x)(x-4)}$$ Where $A(x)$ is our unknown generating function (which we have been looking for). But I don't think this is correct because it does not match the answer down below.
Edit: I have finally figured it out, I will be back with my LaTex code to present it.
Final edit:
Find an explicit formula for $a_n$ using the generation function technique.
\newline We will begin by multiplying each side by $x^n$ and summing over $n\geq1$. We will also let $A(x)=\sum_{n=0}^\infty a_nx^n$ be the unknown generating function which we are looking for.
$$\begin{align*}
\implies \sum_{n=1}^\infty a_nx^n&=\sum_{n=1}^\infty (4a_{n-1}+1)x^n
\end{align*}
$$
Because
$a_0=0$ is given, we can rewrite $A(x)$:
$$
\begin{align*}
A(x)=a_0+\sum_{n=1}^\infty a_nx^n=\sum_{n=1}^\infty a_nx^n
\end{align*}
$$
Now, we continue to manipulate our equality.
$$
\begin{align*}
\implies \sum_{n=1}^\infty a_nx^n&=\sum_{n=1}^\infty (4a_{n-1}+1)x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n + x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n+\sum_{n=1}^\infty x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n+ \frac{x}{1-x} \\[2mm]
&=\sum_{n=0}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=a_0+\sum_{n=1}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=\sum_{n=1}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=4x\sum_{n=1}^\infty a_{n}x^{n}+ \frac{x}{1-x}. \\[2mm]
\end{align*}$$
It is clear now, that we have a multiple of our previously stated $A(x)$ on the left and right hand side.
$$
\begin{align*}
\implies&\sum_{n=1}^\infty a_{n}x^{n}=4x\sum_{n=1}^\infty a_{n}x^{n}+ \frac{x}{1-x} \\[2mm]
\implies&A(x)=4xA(x)+\frac{x}{1-x} \\[2mm]
\implies&A(x)-4xA(x)=\frac{x}{1-x}\\[2mm]
\implies& A(x)(1-4x)=\frac{x}{1-x}\\[2mm]
\implies& A(x)=\frac{x}{(1-x)(1-4x)}=\frac{x}{4x^2-5x+1} \\[2mm]
\end{align*}$$
Thus, the explicit formula for $a_n$ is given by $A(x)=\dfrac{x}{4x^2-5x+1}$ through the use of generating functions.
Finally, it follows that we have $$A(x)=\sum_{n=0}^\infty\frac{4^n-1}{3}x^n=\frac{x}{4x^2-5x+1}.$$
| A way to go about this is to define
$$f(x) = \sum_{n=0}^{\infty} a_n x^n$$
Then
$$x f(x) = \sum_{n=0}^{\infty} a_n x^{n+1} = \sum_{n=1}^{\infty} a_{n-1} x^n$$
Using your recurrence, we find that
$$(1-4 x) f(x) = a_0 + \sum_{n=1}^{\infty} (a_n-4 a_{n-1}) x^n = a_0 + \sum_{n=1}^{\infty} x^n = a_0 + \frac{x}{1-x}$$
Note that $a_0=0$, so that we now have the generating function $f(x)$:
$$f(x) = \frac{x}{(1-x)(1-4 x)} = \frac{x}{1-5 x+4 x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/777808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
The proof of $e^x \leq x + e^{x^2}$ How can we prove the inequality $e^x \le x + e^{x^2}$ for $x\in\mathbb{R}$?
| Note the inequality $e^t \ge t + 1$ for all $t \in \mathbb{R}$.
In particular $e^{x^2} \ge x^2 + 1$.
If $x \le -1$, then
$$
e^{x^2} - e^x + x \ge x^2 + 1 - e^0 + x = x(x+1) \ge 0.
$$
If $-1 < x < 1$, then
\begin{align*}
e^{x^2} - e^x + x
&\ge x^2 + x + 1 - \sum_{k \ge 0} \frac{x^k}{k!} \\
&= \frac{x^2}{2!} - \sum_{k \ge 3} \frac{x^k}{k!} \\
&\ge \frac{x^2}{2!} - \sum_{k \ge 3} \frac{x^2}{k!} \\
&= x^2 \left( \frac12 - \left[e - 1 - \frac{1}{1!}
- \frac{1}{2!} \right]\right) \\
&= x^2 \left( 3 - e \right) \ge 0.
\end{align*}
Finally, if $x \ge 1$, then
$$
e^{x^2} - e^x + x
> e^{x} - e^x + x > 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/781658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
How find the equation $\cot x=\frac{\sin 20^\circ - \sin 80^\circ \cos 20^\circ}{\sin 80^\circ \sin 20^\circ}$ let $x\in R$, and such
$$\cot x =\frac{\sin 20^\circ -\sin 80^\circ \cos 20^\circ}{\sin 80^\circ \sin 20^\circ}$$
Find $x$
my idea:
$$\cot x=\csc 80^\circ - \cot 20^\circ$$
then I can't
| $$\sin80^\circ=2\sin40^\circ\cos40^\circ=2(2\sin20^\circ\cos20^\circ)\cos40^\circ$$
$$\implies\csc80^\circ-\cot20^\circ=\frac{1-4\cos^220^\circ\cos40^\circ}{\sin80^\circ}$$
Now $\displaystyle N= 1-4\cos^220^\circ\cos40^\circ=1-2\cos40^\circ(2\cos^220^\circ)$
Using Double angle formula $\cos2A=2\cos^2A-1,$
$\displaystyle N=1-2\cos40^\circ(1+\cos40^\circ)=1-2\cos40^\circ-2\cos^240^\circ=-2\cos40^\circ-(2\cos^240^\circ-1)$
$\displaystyle=-\cos40^\circ-\underbrace{(\cos40^\circ+\cos80^\circ)}$
Using Prosthaphaeresis Formulas on the under-braced part,
$\displaystyle N=-\cos40^\circ-2\cos60^\circ\cos20^\circ=-(\cos40^\circ+\cos20^\circ)$
Again, $\displaystyle\cos40^\circ+\cos20^\circ=2\cos10^\circ\cos30^\circ$
$$\implies\frac{1-4\cos^220^\circ\cos40^\circ}{\sin80^\circ}=-\frac{2\cos10^\circ\cos30^\circ}{\sin80^\circ}=-2\cos30^\circ=-\sqrt3$$
Actually, the problem came into being as
$$\cot20^\circ-\cot30^\circ=\frac{\cos20^\circ}{\sin20^\circ}-\frac{\cos30^\circ}{\sin30^\circ}=\frac{\sin(30^\circ-20^\circ)}{\sin30^\circ\sin20^\circ}$$
$$=\frac{\sin10^\circ}{\frac12\cdot2\sin10^\circ\cos10^\circ}=\frac1{\cos10^\circ}=\frac1{\sin80^\circ}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/782921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Do the differences of perfect squares apply to perfect cubes and more? I'm curious about a special property of squares. The difference between perfect squares starting from 0 are 1,3,5,7,9..., where each difference goes up by 2. I want to know if there are any patterns for perfect cubes or quartics. Are the differences in a pattern of some sort?
| For squares:
$$1$$
$$\begin{array}{cc}
\color{red}{2} & \color{red}{2} \\
1 & \color{red}{2}
\end{array}$$
$$\begin{array}{ccc}
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} \\
\color{red}{2} & \color{red}{2} & \color{blue}{3} \\
1 & \color{red}{2} & \color{blue}{3}
\end{array}$$
$$\begin{array}{cccc}
\color{green}{4} & \color{green}{4} & \color{green}{4} & \color{green}{4} \\
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} & \color{green}{4} \\
\color{red}{2} & \color{red}{2} & \color{blue}{3} & \color{green}{4} \\
1 & \color{red}{2} & \color{blue}{3} & \color{green}{4}
\end{array}$$
$$\vdots$$
For cubes, where each array is a slice of the cube:
$$1$$
$$\left.\begin{array}{cc}
\color{red}{2} & \color{red}{2} \\
1 & \color{red}{2}
\end{array}\ \ \right| \ \
\begin{array}{cc}
\color{red}{2} & \color{red}{2} \\
\color{red}{2} & \color{red}{2}
\end{array}$$
$$\left.\begin{array}{ccc}
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} \\
\color{red}{2} & \color{red}{2} & \color{blue}{3} \\
1 & \color{red}{2} & \color{blue}{3}
\end{array}\ \ \right| \ \
\left.
\begin{array}{ccc}
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} \\
\color{red}{2} & \color{red}{2} & \color{blue}{3} \\
\color{red}{2} & \color{red}{2} & \color{blue}{3}
\end{array}\ \ \right| \ \
\begin{array}{ccc}
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} \\
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} \\
\color{blue}{3} & \color{blue}{3} & \color{blue}{3}
\end{array}$$
$$\left.\begin{array}{cccc}
\color{green}{4} & \color{green}{4} & \color{green}{4} & \color{green}{4} \\
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} & \color{green}{4} \\
\color{red}{2} & \color{red}{2} & \color{blue}{3} & \color{green}{4} \\
1 & \color{red}{2} & \color{blue}{3} & \color{green}{4}
\end{array} \ \ \right| \ \
\left.
\begin{array}{cccc}
\color{green}{4} & \color{green}{4} & \color{green}{4} & \color{green}{4} \\
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} & \color{green}{4} \\
\color{red}{2} & \color{red}{2} & \color{blue}{3} & \color{green}{4} \\
\color{red}{2} & \color{red}{2} & \color{blue}{3} & \color{green}{4}
\end{array}
\ \ \right| \ \
\left.
\begin{array}{cccc}
\color{green}{4} & \color{green}{4} & \color{green}{4} & \color{green}{4} \\
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} & \color{green}{4} \\
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} & \color{green}{4} \\
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} & \color{green}{4}
\end{array}
\ \ \right| \ \
\begin{array}{cccc}
\color{green}{4} & \color{green}{4} & \color{green}{4} & \color{green}{4} \\
\color{green}{4} & \color{green}{4} & \color{green}{4} & \color{green}{4} \\
\color{green}{4} & \color{green}{4} & \color{green}{4} & \color{green}{4} \\
\color{green}{4} & \color{green}{4} & \color{green}{4} & \color{green}{4}
\end{array}
$$
$$\vdots$$
Count the number of elements of each color.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/786977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Algebraic maximum and minimum based on a constraint Suppose $a,b,c$ are real numbers such that $a^2b^2 + b^2c^2 + c^2a^2 = k$, where $k$ is a constant.
Then the set of all possible values of $abc(a+b+c)$ is?
I attempted writing the constraint in the form of $(abc)^2(1/a + 1/b + 1/c)$ and then tried using $(a+b+c)(1/a+1/b+1/c) >= 9$, but I think I'm missing something
| *
*It is known that $x^2+y^2+z^2\geq xy+yz+zx$. Note $x=ab,y=bc,z=ca, A=abc(a+b+c)$ it follows that $(ab)^2+(bc)^2+(ca)^2 \geq abc(a+b+c)<=>k\geq A.$
*$\dfrac{k}{2}+A = \dfrac{1}{2}[(ab)^2+(bc)^2+(ca)^2]+ abc(a+b+c)= \dfrac{1}{2}[(ab+bc)^2+(bc+ca)^2+(ca+ab)^2] \geq 0=>A\geq -\dfrac{k}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$(x^n)^2=x^n$ for all $n\geq 1$. $x$ is idemotent ring where $x^2= x$ I am asked to prove the above
I have been trying a lot but my teacher did not like they way I solved it,he did not even tell me what is wrong with it.i was thinking if $n=3$ then $x^{2n}$ will be equal to $x^6$ is equal to $$x\cdot x\cdot x\cdot x\cdot x\cdot x=x^6=x^2\cdot x^2\cdot x^2$$
But we know that $x=x^2$
So $x^{2n}=x\cdot x\cdot x = x^3=x^n$
"............sorry can't answer my question now but is this what u mean
(x^n)^2 = x^n for all n>= 1
now for n = 1, we have
x^2 = x which is true for x a member of r
for n+1 we need to show
(x^n+1)^2 = x^n+1
(x^n+1)^2 = (x^n * x)(x^n * x)
(x^n+1)^2 = (x^n)^2 * x^2
(x^n+1)^2 = x^n * x
(x^n+1)^2 = x^n+1
| Hint: $(x^n)^2=x^{2n}=(x^2)^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $g(x) > 0$ If $f(x)$ is a quadratic expression such that $f(x) > 0,\ x\in\mathbb{R}$ and if $g(x)= f(x) + f'(x) + f''(x)$, then prove that $g(x) > 0, \ x\in\mathbb{R}$.
| Let $f(x) = ax^2 + bx + c$, then $f'(x) = 2ax + b$, and $f''(x) = 2a$. So:
$g(x) = ax^2 + (2a + b)x + 2a + b + c$.
Let $x = -\dfrac{c}{b}$, then $f\left(-\dfrac{c}{b}\right) = a\cdot \dfrac{c^2}{b^2} > 0$ implies that $a > 0$.
So: $\triangle_g = (2a + b)^2 - 4a(2a + b + c) = -4a^2 + (b^2 - 4ac) < 0$, since
$b^2 - 4ac < 0$ because $f(x) > 0$ $\forall x$. Thus: $g(x) > 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Eliminating a parameter when intersecting a manifold with a hyperplane In the Euclidean space $\mathbb R^4$ we look at the intersection of the equations$$x^2 + y^2 = 1 \\ z^2 + w^2 = 1$$
sometimes known as the Clifford torus. This is known to be a 2-dimensional manifold, with global parameterization in 2 parameters, given by:
$$x = \cos(2\pi t_1) \\ y = \sin(2\pi t_1) \\ z = \cos(2\pi t_2) \\ w = \sin(2\pi t_2)$$
Now I want to intersect this with an affine hyperplane, given by the equation $$\vec v\cdot(x,y,z,w)=\gamma$$
($\vec v$ is a unit normal vector and $\gamma\in \mathbb R$ is an affine offset.)
Let's assume the intersection is a 1-dimensional manifold - i.e. the Jacobian matrix has rank $3$ at any point in the intersection. (I can show this happens for almost all $\vec v, \gamma$.)
Questions.
*
*Since the intersection is a manifold, there is a 1-parameter local parameterization. But can we say that there is a 1-parameter global parameterization of the intersection manifold?
*Can we say the parameterization is given by $x = \cos(2\pi t)$, $y = \sin(2\pi t)$ and $z,w$ some implicit functions of $t$?
I want to avoid using the quadratic formula and taking square roots because this requires care of when the value is positive and negative... Ideally, I want to use the implicit function theorem to solve this, but I tried to consider it a few times and it's been pretty confusing.
Also - can elimination theory be used to address this question?
| I think that algebraic methods give better result.
Rewrite this system of equations a little differently.
$\left\{\begin{aligned}&G^2+E^2=K^2\\&D^2+F^2=K^2\end{aligned}\right.$
Solutions of this system are of the form:
$G=2XY$
$E=X^2-Y^2$
$D=2ZR$
$F=Z^2-R^2$
$K=X^2+Y^2=Z^2+R^2$
We need solutions will be taken from the following formulas.
This equation is quite symmetrical so formulas making too much can be written:
So for the equation: $X^2+Y^2=Z^2+R^2$
solution:
$X=a(p^2+s^2)$
$Y=b(p^2+s^2)$
$Z=a(p^2-s^2)+2psb$
$R=2psa+(s^2-p^2)b$
solution:
$X=p^2-2(a-2b)ps+(2a^2-4ab+3b^2)s^2$
$Y=2p^2-4(a-b)ps+(4a^2-6ab+2b^2)s^2$
$Z=2p^2-2(a-2b)ps+2(b^2-a^2)s^2$
$R=p^2-2(3a-2b)ps+(4a^2-8ab+3b^2)s^2$
solution:
$X=p^2+2(a-2b)ps+(10a^2-4ab-5b^2)s^2$
$Y=2p^2+4(a+b)ps+(20a^2-14ab+2b^2)s^2$
$Z=-2p^2+2(a-2b)ps+(22a^2-16ab-2b^2)s^2$
$R=p^2+2(7a-2b)ps+(4a^2+8ab-5b^2)s^2$
solution:
$X=2(a+b)p^2+2(a+b)ps+(5a-4b)s^2$
$Y=2((2a-b)p^2+2(a+b)ps+(5a-b)s^2)$
$Z=2((a+b)p^2+(7a-2b)ps+(a+b)s^2)$
$R=2(b-2a)p^2+2(a+b)ps+(11a-4b)s^2$
solution:
$X=2(b-a)p^2+2(a-b)ps-as^2$
$Y=2((b-2a)p^2+2(a-b)ps+(b-a)s^2)$
$Z=2((b-a)p^2+(3a-2b)ps-(a-b)s^2)$
$R=2(b-2a)p^2+2(a-b)ps+as^2$
solution:
$X=(p^2-s^2)b^2+a^2s^2$
$Y=b^2(p-s)^2-2abs^2+a^2s^2$
$Z=b^2(p-s)^2+2abps-a^2s^2$
$R=s^2(a-b)^2+2abps-p^2b^2$
number $a,b,p,s$ integers and sets us, and may be of any sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/790211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is there an elegant way to simplify $\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}$ I wonder how to solve this equation: $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)$$ in an elegant/shorter way.
My way:
$$\frac{\sin(x+20^{\circ })-\sin(x+20^{\circ })\cdot cos(x+20^{\circ})}{\sin(x+20^{\circ })+\sin(x+20^{\circ })\cdot \cos(x+20^{\circ})}=2[1-(\cos(x+20^{\circ}))]$$
so
$$1-\cos(x+20^{\circ})=2[1-\cos(x+20^{\circ})][1+\cos(x+20^{\circ})]$$
which gives
$$\cos(x+20^{\circ})=1\Rightarrow x=-20^{\circ}+360^{\circ}k\\\cos(x+20^{\circ})=-0.5\Rightarrow x=140^{\circ}+360^{\circ}k,\, \, x=-100^{\circ}+360^{\circ}k$$
Thanks.
| $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=\\\frac{\frac{\sin(x+20^{\circ })}{\cos(x+20^{\circ })}-\sin(x+20^{\circ })}{\frac{\sin(x+20^{\circ })}{\cos(x+20^{\circ })}+\sin(x+20^{\circ })}=\\\frac{\sin(x+20^{\circ })}{\sin(x+20^{\circ })}\frac{\frac{1}{\cos(x+20^{\circ })}-1}{\frac{1}{\cos(x+20^{\circ })}+1}=\\\frac{\sin(x+20^{\circ })}{\sin(x+20^{\circ })}\frac{\cos(x+20^{\circ })}{\cos(x+20^{\circ })}\frac{1-\cos(x+20^{\circ })}{1+\cos(x+20^{\circ })}=\\\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\frac{2\sin^2(\frac{x}{2}+10^{\circ })}{2\cos^2(\frac{x}{2}+10^{\circ })}=\\\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\tan^2\left(\frac{x}{2}+10^{\circ }\right)$$ Thus you can rewrite $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)$$ as $$\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\tan^2\left(\frac{x}{2}+10^{\circ }\right)=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)\\\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\frac{1}{\cos^2(\frac{x}{2}+10^{\circ })}=\frac{4}{\cos^2(\frac{x}{2}+10^{\circ })}\\\frac{\tan(x+20^{\circ })}{\cos^2(\frac{x}{2}+10^{\circ })}=\frac{4\tan(x+20^{\circ })}{\cos^2(\frac{x}{2}+10^{\circ })}\\\frac{3\tan(x+20^{\circ })}{\cos^2(\frac{x}{2}+10^{\circ })}=0\Rightarrow\\\tan(x+20^{\circ })=0\\x+20^{\circ }=0^{\circ }\\x=-20^{\circ }+k\pi^{\circ }\qquad k\in\Bbb{Z}$$
| {
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"url": "https://math.stackexchange.com/questions/790632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Complex Numbers Exercise If $a,b,c$ are complex numbers with $a+b+c=0$ and $\|a\|=\|b\|=\|c\| = r>0$ then prove that $$a^{2^n} + b^{2^n} + c^{2^n} = 0$$
Any ideas?
Thanks!
| We have
$$
a + b + c = 0
$$
Taking the conjugate and multiplying by $abc$,
$$
abc \overline{a} + abc \overline{b} + abc \overline{c} = 0
$$
i.e.
$$
\newcommand{\abs}[1]{\left| #1 \right|}
bc \abs{a}^2 + ac \abs{b}^2 + ab \abs{c}^2
$$
But then $\abs{a} = \abs{b} = \abs{c}$, so
$$
ab + bc + ca = 0
$$
It follows that
$$
a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 0
$$
In summary, we have shown that if $a + b + c = 0$ and $a, b, c$ have the same norm, $a^2 + b^2 + c^2 = 0$.
Since $a^2, b^2, c^2$ again have the same norm, it follows by induction that
$$
a^{2^n} + b^{2^n} + c^{2^n} = 0
$$
for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/793702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find bounded solutions of this ODE The function $y(x)$ satisfies the following ode:
$L[y]=\frac{d}{dx}(x^2\frac{dy}{dx})-x^2y= f(x)$
By substituting $y(x)=\frac{z(x)}{x}$ we can solve the homogeneous problem $(L[y]=0)$ and obtain the answer
$y(x)=\frac{Ae^x}{x}+\frac{Be^{-x}}{x}$.
How would you show that the solution given by $sinh(x)/x$ is bounded as $x \rightarrow 0$?
Furthermore how would you find a solution that remains bounded as $x \rightarrow \infty$?
| Hint on the first question:
$$f(x) = \frac{\sinh(x)}{x} = \frac{1}{2x}(e^x - e^{-x})$$
We can show that this function has limit at $x=0$ as follows:
$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$
$e^{-x} = \sum_{n=0}^\infty \frac{(-1)^n x^n}{n!}$
\begin{align*}\frac{\sinh(x)}{x} &= \frac{1}{2x} \left(\sum_{n=0}^\infty \frac{x^n}{n!} - \sum_{n=0}^\infty \frac{(-1)^nx^n}{n!}\right)\\
&= \frac{1}{2x} \sum_{n=0}^\infty (1-(-1)^n)\frac{x^n}{n!}\\
&= \frac{1}{2x} \sum_{n=0}^\infty \frac{2 x^{2n+1}}{(2n+1)!}\\
&=\sum_{n=0}^\infty \frac{x^{2n}}{(2n+1)!}
\end{align*}
So we conclude that
$$\lim_{x\to 0^+} \frac{\sinh(x)}{x} = \lim_{x\to 0^+} \sum_{n=0}^\infty \frac{x^{2n}}{(2n+1)!} = 1$$
This tells us that $\lim_{x\to 0^+} f(x) = 1$, hence the function remains bounded as $x \to 0$.
Hint on the second question:
$$\lim_{x\to\infty} \frac{e^x}{x} = +\infty$$
What this means is that any nonzero constant in front of the term $\frac{e^x}{x}$ will result in an unbounded solution as $x\to\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/797852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Algebraic solution for the intersection point(s) of two parabolas I recently ran through an algebraic solution for the intersection point(s) of two parabolas $ax^2 + bx + c$ and $dx^2 + ex + f$ so that I could write a program that solved for them. The math goes like this:
$$
ax^2 - dx^2 + bx - ex + c - f = 0 \\
x^2(a - d) + x(b - e) = f - c \\
x^2(a - d) + x(b - e) + \frac{(b - e)^2}{4(a - d)} = f - c + \frac{(b - e)^2}{4(a - d)} \\
(x\sqrt{a - d} + \frac{b - e}{2\sqrt{a - d}})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\
(a - d)(x + \frac{b - e}{2(a - d)})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\
x + \frac{b - e}{2(a - d)} = \sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} \\
x = \pm\sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} - \frac{b - e}{2(a - d)} \\
$$
Then solving for $y$ is as simple as plugging $x$ into one of the equations.
$$
y = ax^2 + bx + c
$$
Is my solution for $x$ and $y$ correct? Is there a better way to solve for the intersection points?
| Well, you somehow lost the factor of 4, otherwise seems valid. As you are going to write a program for this, you should also write a solution gfor the case $a=d$ and deal with the case the radical is zero or does not exist.
Actually, you did not have to write all this, as people usually know the solution of quadratic equation (isn't it school material?). For $ax^2 + bx +c = 0$ usually _the discriminant _$D=b^2 - 4ac$ is calculated, and then $x_{1,2}$ are given as $\frac {-b ±\sqrt D}{2a}$. If D=0 then the only sulution $-\frac b {2a}$ exists, and if $D<0$ there are no real solutions (as $\sqrt D$ is not a real number then or does not exist if we only work with real numbers).
Then just use $a-d,b-e$ and $c-f$ as coefficients in the general solution formula. A separate solution for the case $a=d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/799519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to prove $\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$ Question:
let $a,b,c>0$,show that:
$$\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$$
maybe this inequality can Use Holder inequality to solve it
$$\left(\sum_{cyc}\sqrt{\dfrac{ab}{2a^2+bc+ca}}\right)^2\sum_{cyc}(2a^2+bc+ca)\ge(\sum_{cyc}\sqrt[3]{ab})^3$$
then I can't prove it
| We can use also, AM-GM, C-S and uvw.
Indeed, $$\sum_{cyc}\sqrt{\frac{ab}{2a^2+bc+ca}}=\sum_{cyc}\frac{4ab}{2\sqrt{4ab(2a^2+bc+ca)}}\geq$$
$$\geq\sum_{cyc}\frac{4ab}{2a^2+bc+ca+4ab}=\sum_{cyc}\frac{4b^2}{\frac{b}{a}(2a^2+bc+ca+4ab)}\geq$$
$$\geq\frac{4(a+b+c)^2}{\sum\limits_{cyc}\frac{b}{a}(2a^2+bc+ca+4ab)}.$$
Thus, it's enough to prove that
$$8(a+b+c)^5\geq81\sum_{cyc}b^2c(2a^2+bc+ca+4ab)$$ or
$$8(a+b+c)^5\geq81\sum_{cyc}(a^3b^2+4a^3bc+3a^2b^2c)$$ or
$$8(a+b+c)^5-81\sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+3a^2b^2c)+81\sum_{cyc}a^3c^2\geq0.$$
But by Rearrangement
$$\sum_{cyc}a^3c^2=a^2b^2c^2\sum_{cyc}\frac{a}{b^2}\geq a^2b^2c^2\sum_{cyc}\left(a\cdot\frac{1}{a^2}\right)=\sum_{cyc}a^2b^2c.$$
Id east, it's enough to prove that
$$8(a+b+c)^5-81\sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+2a^2b^2c)\geq0.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove a linear inequality of $w^3$, which says that it's enough to prove
the last inequality for an extreme value of $w^3,$ which happens in the following cases.
*
*$w^3\rightarrow0^+$.
Let $c\rightarrow0^+$ and $b=1$.
Thus, we need to prove that
$$8(a+1)^5\geq81a^2(a+1),$$ which is true by AM-GM:
$$8(a+1)^5=8(a+1)^4(a+1)\geq8\left(2\sqrt{a}\right)^4(a+1)=128a^2(a+1)>81a^2(a+1);$$
2. $b=c=1$.
We need to prove that
$$8(a+2)^5\geq81(2a^3+2a^2+2+4a^3+8a+4a^2+2a)$$ or
$$(a-1)^2(4a^3+48a^2+9a+47)\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$ Hi how can we prove this integral below?
$$
I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}
$$
I tried to use
$$
I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx
$$
and now tried changing variables to $y=x(1-x)$ in order to write
$$
I\propto \int_0^1 \sum_{n=0}^\infty y^n
$$
however I do not know how to manipulate the $\log^2 x$ term when doing this procedure when doing this substitution. If we can do this the integral would be trivial from here.
Complex methods are okay also, if you want to use this method we have complex roots at $x=(-1)^{1/3}$. But what contour can we use suitable for the $\log^2x $ term?
Thanks
| Consider the integral
\begin{align}
I = \int_{0}^{1} \frac{\ln^{2}(x)}{1 - x + x^{2}} \ dx
\end{align}
Now consider the factorization of $x^{2} - x + 1$ which is $(x - a)(x-b)$ where $a$ and
$b$ are $e^{\pi i/3}$ and $e^{-\pi i/3}$, respectively. With this in mind it is seen that
\begin{align}
\frac{1}{x^{2} - x + 1} = \frac{1}{a-b} \left( \frac{1}{x - a} - \frac{1}{x-b} \right).
\end{align}
This can also be expanded into series form and is
\begin{align}
\frac{1}{x^{2} - x + 1} = \frac{1}{a-b} \ \sum_{n=0}^{\infty} \left( - \frac{1}{a^{n+1}}
+ \frac{1}{b^{n+1}} \right) x^{n}.
\end{align}
Now consider the integral
\begin{align}
I_{n} &= \int_{0}^{1} x^{n} \ln^{2}(x) \ dx = \partial_{n}^{2} \int_{0}^{1} x^{n} \ dx \\
&= \partial_{n}^{2} \left( \frac{1}{n+1} \right) \\
&= \frac{2}{(n+1)^{3}}.
\end{align}
Since the components are built the desired integral is seen as the following.
\begin{align}
I &= \int_{0}^{1} \frac{\ln^{2}(x)}{1 - x + x^{2}} \ dx \\
&= \frac{1}{a-b} \ \sum_{n=0}^{\infty} \left( - \frac{1}{a^{n+1}}
+ \frac{1}{b^{n+1}} \right) \ \int_{0}^{1} x^{n} \ln^{2}(x) \ dx \\
&= \frac{1}{a-b} \ \sum_{n=0}^{\infty} \left( - \frac{1}{a^{n+1}}
+ \frac{1}{b^{n+1}} \right) \frac{2}{(n+1)^{3}} \\
&= \frac{2}{a-b} \ \sum_{n=1}^{\infty} \left( - \frac{1}{a^{n}}
+ \frac{1}{b^{n}} \right) \frac{1}{n^{3}} \\
&= \frac{2}{a-b} \ \sum_{n=1}^{\infty} \left( \frac{a^{n}-b^{n}}{(ab)^{n}}
\right) \frac{1}{n^{3}} \\
&= \frac{2}{a-b} \ \sum_{n=1}^{\infty} \left( \frac{a^{n}}{n^{3}}
- \frac{b^{n}}{n^{3}} \right) \\
&= \frac{2}{a-b} \left[ Li_{3} (a) - Li_{3}(b) \right],
\end{align}
where $Li_{3}(x)$ is the trilogarithm. Utilizing the results
\begin{align}
Li_{3}(a) &= Li_{3}(e^{\pi i/3}) = \frac{1}{3} \zeta(3) + \frac{5 \pi^{3} i }{162} \\
Li_{3}(b) &= Li_{3}(e^{-\pi i/3}) = \frac{1}{3} \zeta(3) - \frac{5 \pi^{3} i }{162} \\
a-b &= e^{\pi i /3} - e^{- \pi i/3} = \sqrt{3} i
\end{align}
then
\begin{align}
I &= \frac{2}{\sqrt{3} i} \cdot \frac{5 \pi^{3} i}{81} = \frac{10 \pi^{3}}{81 \sqrt{3}}.
\end{align}
Hence
\begin{align}
\int_{0}^{1} \frac{\ln^{2}(x)}{1 - x + x^{2}} \ dx = \frac{10 \pi^{3}}{81 \sqrt{3}} .
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 8,
"answer_id": 5
} |
Inequality $\sum\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum \frac{1}{x + n^2} $
$x\geq0$, then, we have
$$\sum_{n=1}^{\infty}\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{x + n^2} $$
The problem is not easy, even $x=1$. Any help will be appreciated
| The problem is easy for $ 0 \leq x \leq 1$. We have $x \leq n^2$, and hence $ \frac{ x}{x+n^2} \leq \frac{1}{2}$. Thus,
$$ \sum_{n=1}^{\infty} \frac{ x}{(x+n^2)^2} \leq \sum_{n=1}^{\infty} \frac{1}{2} \times \frac{1}{x+n^2}.$$
It remains to verify that we have strict inequality in at least one case.
It is interesting to note that $\int_0^\infty \frac{ x} { (x+y^2)^2} \, dy =\frac{1}{2} \int_0^\infty \frac{1}{x+y^2} \, dy $. This possibly motivates the analysis approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Do these integrals have a closed form? $I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$ The following integrals look like they might have a closed form, but Mathematica could not find one. Can they be calculated, perhaps by differentiating under the integral sign?
$$I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$$
$$I_2 = \int_{-\infty }^{\infty } \frac{\sin ^2(x)}{x \sinh (x)} \, dx$$
| For the first one we need:
$$\int _{-1/2}^{1/2}\!{{\rm e}^{2\,iax}}{da}={\frac {\sin \left( x
\right) }{x}}\tag{1}$$
$$ \frac{1}{\cosh \left( x \right)}=-2\,\sum _{n=1}^{\infty
} \left( -1 \right) ^{n}{{\rm e}^{- \left| x \right| \left( 2\,n-1
\right) }}\tag{2}$$
$$\int _{-\infty }^{\infty }\!{{\rm e}^{2\,iax}}{{\rm e}^{- \left| x
\right| \left( 2\,n-1 \right) }}{dx}=- \frac{1}{\left( 2\,ia-2\,n+1 \right) }- \frac{1}{\left( -2\,ia-2\,n+1 \right)}\tag{3}$$
$$-2\,\sum _{n=1}^{\infty } \left( -1 \right) ^{n} \left(- \frac{1}{\left( 2\,ia-2\,n+1 \right) }- \frac{1}{\left( -2\,ia-2\,n+1 \right)} \right) ={
\frac {\pi }{\cosh \left( \pi \,a \right) }}\tag{4}$$
we get:
$$
\begin{aligned}
\int _{-\infty }^{\infty }\!{\frac {\sin \left( x \right) }{x\cosh
\left( x \right) }}{dx}&=\int _{-1/2}^{1/2}\!{\frac {\pi }{\cosh
\left( \pi \,a \right) }}{da}\\
&=2\,\arctan \left( {
{\rm e}^{1/2\,\pi }} \right)-2\,\arctan \left( {{\rm e}^{-1/2\,\pi }} \right)\\
&=2\,\arctan \left( \sinh \left( \frac{1}{2}\,\pi \right) \right)
\end{aligned}\tag{5}$$
where the last part follows from $(2)$ and the Taylor series for arctan:
$$\arctan \left( x \right) =\sum _{n=0}^{\infty }{\frac { \left( -1
\right) ^{n}{x}^{2\,n+1}}{2\,n+1}}\tag{6}$$
For the second one we need:
$$ \frac{1}{\sinh \left( x \right) }=2\,\sum _{n=1}^{\infty }
{{\rm e}^{-x \left( 2\,n-1 \right) }}\tag{7}$$
$${\frac { \sin^2 \left( x \right)}{x}}=-\frac{1}{2}\,\sum _{
m=1}^{\infty }{\frac { \left( -1 \right) ^{m}{2}^{2\,m}{x}^{2\,m-1}}{
\left( 2\,m \right) !}}\tag{8}$$
$$\int _{0}^{\infty }\!{x}^{2\,m-1}{{\rm e}^{-x \left( 2\,n-1 \right) }}
{dx}={\frac { \left( 2\,m-1 \right) !}{ \left( 2\,n-1 \right) ^{2\,m}}
}\tag{9}$$
$$\cot \left( z \right) -\frac{1}{z}=-\frac{2}{\pi}\,\sum _{m=1}^{\infty }\zeta
\left( 2\,m \right) \left( {\frac {z}{\pi }} \right) ^{2\,m-1}\tag{10}$$
From $(6,7,8)$:
$$
\begin{aligned}
\int _{0}^{\infty }\!{\frac { \sin^2 \left( x \right)}{x\sinh \left( x \right) }}{dx}&=-\frac{1}{2}\,\sum _{m=1}^{\infty } \left(
\frac{\left( -4 \right) ^{m}}{m}\sum _{n=1}^{\infty }{\frac {1}{\left( 2\,n-1
\right) ^{2\,m}}} \right)\\
&=\frac{1}{4}\sum _{m=1}^{\infty }\,{\frac
{\zeta \left( 2\,m \right) \left( {4}^{m}-1 \right) \left( -1
\right) ^{m}}{m}}
\end{aligned}\tag{11}$$
and after integrating $(10)$ once we know that:
$$\ln \left( {\frac {\sin \left( z \right) }{z}} \right) =-\sum _{m=1}^
{\infty }\frac{\zeta \left( 2\,m \right)}{m} \left( {\frac {z}{\pi }} \right)
^{2\,m}\tag{12}
$$
so by comparing $(11)$ with $(12)$ we know that:
$$
\begin{aligned}\int _{0}^{\infty }\!{\frac { \sin^2 \left( x \right) }{x\sinh \left( x \right) }}{dx}&=\frac{1}{2}\,\ln\!\left( \frac{1}{2}\,{\frac {\sinh
\left( 2\,\pi \right) }{\sinh \left( \pi \right) }} \right)\\
&=\frac{1}{2}\,\ln \left( \cosh \left( \pi
\right) \right)
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/804483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 3
} |
Summation with Ceilinged Logarithmic Function According to Johann Blieberger's paper - "Discrete Loops and Worst Case Performance" (1994):
$$
\sum_{i = 1}^{n}\left \lceil \log_2{(i)} \right \rceil = n\left \lceil \log_2{(n)} \right \rceil - 2^{\left \lceil \log_2{(n)} \right \rceil} + 1
$$
Now, I was wondering if someone knows what the following may equal?
$$
\sum_{i = 1}^{n}i\left \lceil \log_2{(i)} \right \rceil = ?
$$
| I could come with a solution with an incisive precision for any base a by doing this:
$$
\\
T(n) = \sum_{i= 1}^{n}i\left \lceil \log_a(i) \right \rceil = \left[\sum_{i= 1}^{\left \lfloor \log_a(n) \right \rfloor}i \left(\sum_{j = a^{i - 1} + 1}^{a^i} j\right)\right] + \left \lceil \log_a(n) \right \rceil\sum_{i = a^{\left \lfloor \log_a(n) \right \rfloor} + 1}^{n}i
\\
T(n) = \left[\frac{1}{2}\sum_{i= 1}^{\left \lfloor \log_a(n) \right \rfloor}i \left(a^{i - 2}(a -1)(a^{i + 1} + a^i + a)\right)\right] + \left \lceil \log_a(n) \right \rceil\sum_{i = a^{\left \lfloor \log_a(n) \right \rfloor} + 1}^{n}i
\\
\text{Let }
T_1(n) = \left[\frac{1}{2}\sum_{i= 1}^{\left \lfloor \log_a(n) \right \rfloor}i \left(a^{i - 2}(a -1)(a^{i + 1} + a^i + a)\right)\right]
\\
\text{And }
T_2(n) = \left \lceil \log_a(n) \right \rceil\sum_{i = a^{\left \lfloor \log_a(n) \right \rfloor} + 1}^{n}i
\\
T_1(n) = \frac{1}{2(a^2 - 1)}\left((-1 - \left \lfloor \log_a(n) \right \rfloor)a^{2\left \lfloor \log_a(n) \right \rfloor} - a^{\left \lfloor \log_a(n) \right \rfloor + 1} + \left \lfloor \log_a(n) \right \rfloor a^{\left \lfloor \log_a(n) \right \rfloor+ 2} + \left \lfloor \log_a(n) \right \rfloor a^{2\left \lfloor \log_a(n) \right \rfloor+ 2} - (\left \lfloor \log_a(n) \right \rfloor + 1)a^{\left \lfloor \log_a(n) \right \rfloor} + a + 2 \right)
\\
T_2(n) = \frac{\left \lceil \log_a(n) \right \rceil(n - a^{\left \lfloor \log_a(n) \right \rfloor})(a^{\left \lfloor \log_a(n) \right \rfloor} + n + 1)}{2}
\\\\
\text{Therefore }
T(n) = T_1(n) + T_2(n)
$$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$ \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ$ How can I find the following product using elementary trigonometry?
$$ \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ.$$
I have tried using a substitution, but nothing has worked.
| First, let's re-arrange these terms so that we can make use of the hints in other answers.
$$\tan(1^\circ) \cdot \tan(89^\circ) \cdot \tan(2^\circ) \cdot \tan(88^\circ) \cdot\cdot\cdot \tan(44^\circ) \cdot \tan(46^\circ) \cdot \tan(45^\circ)$$
Here, we can see a clear pattern of $$\tan(x) \cdot\tan(90^\circ-x)$$ repeating, except for 45, who has no dance partner.
Now, we can use the fact that $$\tan(x) \cdot \tan(90^\circ - x) = 1$$ and reduce all the pairs of numbers to 1. We're left with
$$ 1 \cdot 1 \cdot 1 \cdot \cdot \cdot \tan(45^\circ)$$
and since $$ tan(45^\circ) = 1 $$
we get an answer of $$\tan(1^\circ) \cdot \tan(2^\circ) \cdot \tan(3^\circ) \cdot\cdot\cdot \tan(89^\circ) = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/805704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
solving a system of equations dealing with Lorentz transformations Can anyone help me to find the solutions of this system of equations:
$$c^2x^2-v^2y^2=c^2$$
$$y^2-c^2z^2=1$$
$$vy^2+c^2zx=0$$
I know the answer:
$$x= \frac{1}{ \sqrt{1- \frac{ v^{2} }{ c^{2} } } } $$
$$y= \frac{1}{ \sqrt{1- \frac{ v^{2} }{ c^{2} } } } $$
$$z= -\frac{v}{c^2} \sqrt{1- \frac{ v^{2} }{ c^{2} } } $$
But I can't follow the steps. If any one could say me how to solve this kind of problem that would be helpful for me. Thank you in advance.
| $$c^2x^2-v^2y^2=c^2$$
$$y^2-c^2z^2=1$$
$$vy^2+c^2zx=0$$
mutliplying the second by $x^{2}$ we find
$$
y^2x^2 - c^2z^2x^2 = x^2
$$
using the third equation we find
$$
y^2x^2 - c^2z^2x^2 = y^2x^2 - c^2\left(\frac{vy^2}{c^2}\right)^2 = x^2
$$
or
$$
y^2x^2 - \frac{v^2}{c^2}y^4 = x^2
$$
now we can use the first equation to yield
$$
x^2y^2 = \frac{v^2y^4 + c^2y^2}{c^2}
$$
and subbing into the previous equation we obtain
$$
v^2y^4 + c^2y^2 - v^2y^4 = c^2 + v^2y^2
$$
or
$$
y^2 = \frac{c^2}{c^2-v^2} = \frac{1}{1-\left(\frac{v}{c}\right)^2}
$$
now you have y, you immediately obtain the remaining results.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve the following? $ x^3+1=2{(2x-1)}^{1/3} $. Find all the real solutions of
$$x^3+1=2{(2x-1)}^{1/3} $$
I tried to cube both sides but got messed up with a nine degree equation! Please help.
Thanks in advance!
| We are given
$$x^3+1=2{(2x-1)}^\frac{1}{3}$$
Substitute $(2x-1)=y$
$\implies x^3=2y-1$
Also,
$y^3=2x-1$ [Using the initial substitution]
Subtracting the above two equations, we get,
$x^3-y^3=-2(x-y)$
$\implies x=y$
or,
$x^2+xy+y^2=-2$
But since $x^2+y^2 \ge xy$ and L.H.S. is thus positive,
$\implies x^2+y^2+xy=-2$ yields no solution.
Now, as $x=y$,
$\implies x=(2x-1)^{\frac{1}{3}}$
$\implies x^3-2x+1=0$
Now, using rational root theorem, $x=1$ satisfies the above equation. So, by dividing with $(x-1)$, we obtain
$(x-1)(x^2+x-1)=0$
$\implies x=1,\frac{-1 \pm \sqrt{5}}{2}$ are the required solutions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Functions and inequalities I have no idea about this question. Please give a hand whoever can. I am a beginner.
Let $f(x)= x^2 - 10$.
Given $\epsilon > 0$, find the value "$a$" so that $$|x -3 |< \min\{2,a\} \implies |f(x)-10|<\epsilon.$$
| Assume that $x \neq 3$ to begin with:
$|f(x) - 10| = |x^2 - 10 - 10| = |x^2 - 20| < \epsilon \iff 20 - \epsilon < x^2 < 20 + \epsilon$.
Now let $t = x - 3$, then $x = t + 3$. So:
$20 - \epsilon < (t+3)^2 < 20 + \epsilon$. Thus:
$\sqrt{20 - \epsilon} < |t+3| < \sqrt{20 + \epsilon}$. Thus:
$-\sqrt{20 + \epsilon} < t + 3 < \sqrt{20 + \epsilon}$, and:
$t + 3 < -\sqrt{20 - \epsilon}$ or $t + 3 > \sqrt{20 - \epsilon}$. So:
$-3 - \sqrt{20 + \epsilon} < t < -3 + \sqrt{20 + \epsilon}$, and:
$t < -3 - \sqrt{20 - \epsilon}$ or $t > -3 + \sqrt{20 - \epsilon}$. Thus:
$-3 - \sqrt{20 + \epsilon} < t < -3 - \sqrt{20 - \epsilon}$ or $-3 + \sqrt{20 - \epsilon} < t < -3 + \sqrt{20 + \epsilon}$. So to give one such "positive" $a$ ( $a > 0$ ), we can take:
$a = \text{min} \{-3 + \sqrt{20 + \epsilon},3 - \sqrt{20 - \epsilon}\}$.
Note that it is possible for $a$ to be positive if you have a big enough $\epsilon$.
| {
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Set of linear equations Find eigenvalues and eigenvectors of the matrix:
$\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix}$
$\begin{pmatrix} 1-\lambda & 0 & -2 \\ 1 & 3-\lambda & -1 \\ -1 & 0 & 2-\lambda \end{pmatrix}$
I have found the eigenvalues, which are: 0 and 3.
For the 0-eig I have found their corresponding eigenvector (-6,1,-3)
and now I try to find the eigenvector for eig = 3:
$\begin{pmatrix} -2 & 0 & -2 \\ 1 & 0 & -1 \\ -1 & 0 & -1 \end{pmatrix}$~$\begin{pmatrix} 1 & 0 & -1 \\ 0 & 0 & 1 \end{pmatrix}$
and I am then supposed to find this solution: (0,1,0)
First, I don't understand how to get from the first matrix to the second matrix.
Secondly, how to find the solution.
May you help me please.
Inverse matrix:
First step:
$$
\left[
\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
-1&2&1&0&1&0\\
0&0&1&0&0&1\\
\end{array}
\right]
$$
Second step:
$$\left[
\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
0&2&1&1&1&0\\
0&0&1&0&0&1\\
\end{array}
\right]
$$
Last step:
$$\left[
\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
0&1&0&1/2&1/2&-1/2\\
0&0&1&0&0&1\\
\end{array}
\right]
$$
| I would gladly write a comment but I have no enough rep. yet.
It seems that your matrix (3x3) has only 2 eigenvalues which means that you can not find the third eigenvector. You have to find a generalized eigenvector, which seems to be your solution.
Regarding your second matrix, I believe that eig = 0 will give you (1 0 -1)' (its corresponding eigenvector transposed), and eig = 3 will give you (0 0 1)'
Again, apologize for using the answer window!
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Prove $\int\frac{12x\sin^{-1}x}{9x^4+6x^2+1}dx=-\frac{2\sin^{-1}x}{3x^2+1}+\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$ How to prove
$$\int\frac{12x\sin^{-1}x}{9x^4+6x^2+1}dx=-\frac{2\sin^{-1}x}{3x^2+1}+\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$$
where $\sin^{-1}x$ and $\tan^{-1}x$ are inverse of trig functions. I don't know how to find the integral because of inverse of trig functions. I missed calc class twice. Please help me. Thanks.
| $$\int \frac{12\sin^{-1} x}{(3x^2+1)^2}\,dx=\sin^{-1}x\int \frac{12x}{(3x^2+1)^2}\,dx -\int \left(\frac{1}{\sqrt{1-x^2}}\int \frac{12x}{(3x^2+1)^2}\,dx\right)\,dx$$
To evaluate $\displaystyle \int \frac{12x}{(3x^2+1)^2}$, use the substitution $3x^2+1=u \Rightarrow 6x\,dx=du$ to get: $\frac{-2}{3x^2+1}$, i.e
$$\int \frac{12\sin^{-1} x}{(3x^2+1)^2}\,dx=\frac{-2\sin^{-1}x}{3x^2+1}+2\int \frac{1}{\sqrt{1-x^2}(3x^2+1)}\,dx$$
To evaluate the last integral, use the substitution $x=\sin\theta \Rightarrow dx=\cos\theta d\,\theta$ to get:
$$2\int \frac{1}{\sqrt{1-x^2}(3x^2+1)}\,dx=2\int \frac{1}{3\sin^2\theta+1}\,d\theta=2\int \frac{\sec^2\theta}{4\tan^2\theta+1}\,d\theta$$
$$\Rightarrow 2\int \frac{\sec^2\theta}{4\tan^2\theta+1}\,d\theta=\tan^{-1}(2\tan\theta)+C$$
Since $\sin\theta=x$, it is easy to see that $\tan\theta=x/\sqrt{1-x^2}$, hence
$$\tan^{-1}(2\tan\theta)=\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$$
Putting everything together, the final answer is:
$$\frac{-2\sin^{-1}x}{3x^2+1}+\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$$
| {
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Reciprocal Identities Help. $\displaystyle\frac{\sin^2A + 2\cos A - 1}{2 + \cos A - \cos^2A}$ = $\displaystyle\frac{1}{\sec(A)+1}$
Can someone help me with this one? I can't seem to get it right, I get lost, right side has me stumped. Sorry; there's the right side.
| Use the identity $\sin^2 x = 1 - \cos^2 x$. Then, factor the left hand side:
$$\frac{\sin^2A + 2\cos A - 1}{2 + \cos A - \cos^2A} = \frac{1 - \cos^2 A+ 2 \cos A - 1}{(2 - \cos A)(1 + \cos A)} = \frac{\cos A(2 - \cos A)}{(2-\cos A)(1 + \cos A)}$$
Now cancel the common factor, to get $$\frac{\cos A}{1 + \cos A}$$
Now, if we divide the numerator and denominator by $\cos A$, the result is $$\frac 1{\frac 1{\cos A} + 1} = \frac 1{\sec A + 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819411",
"timestamp": "2023-03-29T00:00:00",
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If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4,$ then $(a-b)^2=\;$? If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4$, what is the value of $(a-b)^2$? I think $a^2+b^2=36$, please confirm and is it possible to to figure out one of the variables?
| Hints:
$\frac{1}{a^2}+\frac{1}{b^2} = 4 \rightarrow b^2 + a^2 = 4(ab)^2$
$(a-b)^2 = (a^2+b^2) -2(ab)$
edit:
To solve for a particular variable, you can use $ab=3 \rightarrow a = \frac{3}{b}$ to eliminate a variable. For example $a^2+b^2 = \frac{9}{b^2}+b^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $ a^2 b^2 \left( {a^2 + b^2 - 2} \right) \ge \left( {a + b} \right)\left( {ab - 1} \right) $ good evening I want to show that
$(1)a;b\in\mathbb {R^*_+}:a^2 b^2 \left( {a^2 + b^2 - 2} \right) \ge \left( {a + b} \right)\left( {ab - 1} \right)
$
$
\begin{array}{l}
\frac{a}{b} + \frac{b}{a} \ge 2 \\
\frac{a}{b} + \frac{b}{a} = \frac{{a^2 + b^2 }}{{ab}} \Rightarrow a^2 + b^2 = ab\left( {\frac{a}{b} + \frac{b}{a}} \right) \ge 2ab \\
a^2 b^2 \left( {a^2 + b^2 - 2} \right) \ge a^2 b^2 \left( {2ab - 2} \right) = 2a^2 b^2 \left( {ab - 1} \right) \\
\end{array}
$
$ x;y;z\in\mathbb {R^*_+}:$
$(2)\sum_{cyc}^{ } xy(x+y-z)\ge \sqrt {3(x^3y^3+y^3z^3+z^3x^3)}$
thank you in advance
| By way of simplifying the situation, let x = ab and y = a+b. Note that as a and b range over all positive reals, x and y range over all pairs for which x>0 and y > 2 √x because if x lies on the hyperbola ab = x in the first quadrant of the a-b plane, then the least value for a+b will be when a=b and then it is 2 √x. So as long as y > 2 √x there will be positive a and b satisfying the equations between x,y and a,b.
Expressing the given inequality in terms of x and y we get:
x^2 (y^2 – 2x – 2) > y (x-1), which can be transformed into
y^2 + y(1-x)/x^2 > 2x+2.
Let u = (1-x)/2x^2 and v = 2x+2, so we require that
y^2 + 2uy > v, which holds iff y > √(u^2 + v) – u (since y>0).
The only way for this condition to hold for all x, y as described above is for
2 √x >= √(u^2 + v) – u.
Move the u to the other side and square to get
u √x >= v/4 – x.
Now replace u and v with their definitions to get the condition:
(1-x)/2x^2 * √x >= (1+x)/2 – x = (1-x)/2.
If 1-x > 0 this simplifies to x<=1 which is true and if 1-x <= 0 it simplifies to x>=1, also true, which proves the inequality.
To clarify things a little, one starts with an arbitrary positive a and b, then finds x, y, u, v. Since y > 2 √x which itself is >= √(u^2 + v) – u, also y > √(u^2 + v) – u. This last inequality then can be traced back to demonstrating the original inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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Gram-Schmidt process when metric is not an Euclidean Can we use a Gram-Schmidt process for finding an orthogonal vector if a metric on a space is not an Euclidean one?
These are the vectors and the metric.
If we have the next two vectors:
\begin{align}
\notag
A_{1}=c_{1}y_{1}^{2}y_{2}^{2}u(u+2t)(u+y_{2}^{2})\frac{\partial}{\partial x_{2}},
\end{align}
\begin{align}
\notag
A=-c_{11}\frac{\partial}{\partial x_{1}}-c_{12}\frac{\partial}{\partial x_{2}}-\frac{u(u+2t)}{u+y_{2}^{2}}(c_{8}(u+y_{1}^{2})+c_{1}y_{1}y_{2})\frac{\partial}{\partial y_{1}},
\end{align}
and the metric
\begin{align}
\notag
G=
\begin{pmatrix}
u+vy_{1}^{2} & vy_{1}y_{2} & 0& 0 \\
vy_{1}y_{2} & u+vy_{2}^{2} & 0& 0 \\
0 & 0 & \frac{u+vy_{2}^{2}}{u(u+2tv)}& -\frac{vy_{1}y_{2}}{u(u+2tv)}\\
0& 0& -\frac{vy_{1}y_{2}}{u(u+2tv)} & \frac{u+vy_{1}^{2}}{u(u+2tv)} ,
\end{pmatrix}
\end{align}
If we use Gram-Schmidt in this case, we get a vector which is orthogonal to $A_{1}$ but not to $A$.\
On the other hand, we can find vectors that span a normal bundle of a submanifold generated by $A_{1}$ and $A$, using an almost complex structure.\
We need to find a normal component of a given vector with respect to $A_{1}$ and $A$, that is why we need a vector that is orthogonal to $A_{1}$ and $A$. So, my other question would be is one of the vectors in the normal bundle enough for finding mentioned component?
Thank you!
| In the exact same way, but instead of using the dot product, you use whatever inner product you're given.
Example: Consider $\mathbb{R}^2$ with the inner product
$$\langle (x_1, x_2), (y_1, y_2)\rangle = 2x_1y_1 + x_1y_2 + x_2y_1 + 3x_2y_2.$$
Given the basis $\{(1, 0), (0, 1)\}$ of $\mathbb{R}^2$, what do we get when we apply the Gram-Schmidt process? As
$$\|(1, 0)\|^2 = \langle (1, 0), (1, 0)\rangle = 2(1)(1) + (1)(0) + (0)(1) + 3(0)(0) = 2,$$
$u_1 = \frac{1}{\sqrt{2}}(1, 0)$. Now
\begin{align*}
u_2' &= (0, 1) - \left\langle (0, 1), \frac{1}{\sqrt{2}}(1, 0)\right\rangle\frac{1}{\sqrt{2}}(1, 0)\\
&= (0, 1) - \left(2(0)\left(\frac{1}{\sqrt{2}}\right)+(0)(0)+(1)\left(\frac{1}{\sqrt{2}}\right)+3(1)(0)\right)\frac{1}{\sqrt{2}}(1,0)\\
&= (0,1)-\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}(1,0)\\
&= (0,1) - \frac{1}{2}(1, 0)\\
&= \left(-\frac{1}{2}, 1\right)
\end{align*}
and
\begin{align*}
\|u_2'\|^2 &= \left\|\left(-\frac{1}{2},1\right)\right\|^2\\
&= \left\langle\left(-\frac{1}{2},1\right), \left(-\frac{1}{2},1\right)\right\rangle\\
&= 2\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) + \left(-\frac{1}{2}\right)(1) + (1)\left(-\frac{1}{2}\right)+3(1)(1)\\
&= \frac{5}{2}
\end{align*}
so $u_2 = \frac{\sqrt{2}}{\sqrt{5}}(-\frac{1}{2}, 1)$. So an orthonormal basis for $\mathbb{R}^2$, with the inner product above is
$$\{u_1, u_2\} = \left\{\frac{1}{\sqrt{2}}(1, 0), \frac{\sqrt{2}}{\sqrt{5}}\left(-\frac{1}{2}, 1\right)\right\}.$$
| {
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Improper integral comparison test Having this integral
$$\int_1^{\infty}\frac{3x^2+2x +1}{x^3+6x^2+x+4}$$
In order to do the comparison test at some point it gets like
$$\frac{3x^2+2x +1}{x^3+6x^2+x+4}\geq \frac{1}{4x}$$
How is $\frac{1}{4x}$ found ? It doesnt seem obvious to me.
EDIT
Also how is this found
$$0\leq\frac{3x +1}{\sqrt{x^5+4x^3+5x}}\leq \frac{4}{x^{3/2}}$$
Where does 4 came from here ?Minimum value ? 3x1 +1 ? Why.
| For positive $x$, the top is $\ge 3x^2$. For $x\ge 1$, the bottom is $\le x^3+6x^3+x^3+4x^3$.
Edit: For your added question, the range of values of $x$ is not specified. However, if $x\ge 1$, then the top is $\le 4x$. The bottom is $\ge \sqrt{x^5}$. So for $x\ge 1$, the whole thing is $\le \frac{4x}{x^{5/2}}$, which simplifies to $\frac{4}{x^{3/2}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate the inverse of sin How does one integrate $\int\ {\sin^{-1}(x)}$, using integration by parts, where:
$$ u={\sin^{-1}}, du=\frac{1}{\sqrt{1-x^2}},dv=dx, v=x ? $$
This is a partial solution, and I do not quite understand how to move forward from here$$\int\ {\sin^{-1}(x)}= x\sin^{-1} - \int\left( \frac{x}{\sqrt{1-x^2}} \right) dx$$
How do I integrate: $$ \int\left( \frac{x}{\sqrt{1-x^2}} \right) dx?$$
| Let $$ J= \int\left( \frac{x}{\sqrt{1-x^2}} \right) dx.$$
Then $$ J=\int\left( \color{green}{x} \cdot \color{red}{\frac{1}{\sqrt{1-x^2}}} \right) \color{green}{dx}.$$
Let $t=1-x^2 \implies \frac{dt}{dx}=-2x \iff \color{green}{xdx}= -\frac{1}{2}dt$.
Then $J=-\frac{1}{2}\int \color{red}{\frac{1}{\sqrt{t}}}dt=-\frac{1}{2} \int t^{-1/2}dt=-t^{1/2}+C=\underbrace{-\sqrt{t}+C=\boxed{-\sqrt{1-x^2}+C}}_{\text{since} \ t:=1-x^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find positive integers $(x,n)$ such that $x^{n} + 2^{n} + 1$ is a divisor of $x^{n+1} +2^{n+1} +1$ Find all positive integers $(x,n)$ such that $x^{n} + 2^{n} + 1$ is a divisor of $x^{n+1} +2^{n+1} +1$
I encountered this question in one of my monthly assignments. Unfortunately, I don't know how to proceed about this question at all. Please help.
Thanks in advance!
| Check that $n=1$ gives two solutions $x=4$ and $x=11$. From now on $n>1$.
For each single case $x=1$, $x=2$ check that there is no solution.
Now we will consider
$$
x(x^n+2^n+1)-(x^{n+1}+2^{n+1}+1)=2^n(x-2)+x-1
$$
instead of $x^{n+1}+2^{n+1}+1$.
Check that $x=3$ gives no solution. From now on $x>3$ and $n>1$, hence
$$
x^{n-1}(x-2)\geq 2^n(x-2),
$$
$$
x^{n-1}\cdot 2\geq x-1,
$$
$$
2^n+1>0.
$$
Summing last three lines we get
$$
x^n+2^n+1>2^n(x-2)+x-1
$$
and the left hand side is not a divisor of the right hand side, there is no solution for $n>1$, $x>3$.
| {
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Two ways to show that $\sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ Show that: $\large \sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ on: $0<x<\frac {\pi}2$
I tried to solve it in two ways and got a little stuck:
One way is to use Cauchy's MVT, define $f,g$ such that $f(x)=\sin x -x +\frac {x^3}{3!}$ and $g(x)=\frac {x^5}{5!}$ so $\frac {f(x)}{g(x)}<1$ and both are continuous and differentiable on $x\in (0,\frac{\pi}2)$ so
$$\large \frac {f(x)}{g(x)}=\frac{f(x)-f(0)}{g(x)-g(0)}=\frac {f'(c)}{g'(c)}\Rightarrow
\frac{\sin x -x +\frac {x^3}{3!}} {\frac {x^5}{5!}}=\frac {\cos y -1 +\frac{y^2}{2}}{\frac{y^4}{24}}$$
Such that $y\in (0,x)$.
Now what ? I need to show that $\frac {f'(y)}{g'(y)}<1$ but I tried to input $\frac{\pi} 2$ as a value to $f'/g'$ but it was larger than 1.
The other way: With Taylor expansion.
$\large \sin x=x-\frac {x^3}{3!}+\frac {x^5}{5!}+R_{5,0}(x)\Rightarrow\\
\large \sin x-x+\frac {x^3}{3!}-\frac {x^5}{5!}=R_{5,0}(x)$
Now we're left with showing that $R_{5,0}(x)<0$
So $\large R_{5,0}(x)=\frac{-\sin(c)c^7}{7!}$ for $c\in (0,x)$, now can I say that because it's negative and $(\sin(c)c^7>0)$ for all $c\in(0,x)$, $R_{5,0}(x)<0$ ?
Also, would it be correct to say that: $\large R_{5,0}(x)=\frac{cos(c)c^6}{6!}$?
| Use increasing functions.
For example
$$ 1- \frac{x^2}{2} <\cos x$$
take
$$f(x)=\cos x +\frac{x^2}{2}-1$$ then
$$f^{\prime}=-\sin x + x >0$$ so $f$ is increasing, and $f(0) <f(x)$ gives the desired inequality. Next take $x-\frac{x^3}{3} < \sin x$ then go back to the corresponding $cos$ expansion and finally your ready to prove your inequality. Each step relies on the last.
| {
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Finding an equation of circle which passes through three points How to find the equation of a circle which passes through these points $(5,10), (-5,0),(9,-6)$
using the formula
$(x-q)^2 + (y-p)^2 = r^2$.
I know i need to use that formula but have no idea how to start, I have tried to start but don't think my answer is right.
| $\begin{vmatrix}
x^2+y^2&x&y&1\\
5^2+10^2&5&10&1\\
(-5)^2+0^2&-5&0&1\\
9^2+(-6)^2&9&-6&1\\
\end{vmatrix}=
\begin{vmatrix}
x^2+y^2&x&y&1\\
125&5&10&1\\
25&-5&0&1\\
117&9&-6&1\\
\end{vmatrix}
=
0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Solving a simple matrix polynomial Does there exist a $2\times 2$ Matrix $A$ such that
$A-A^2=\begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$ ?
| Assume such an $A$ exists. Let $\lambda_1$ and $\lambda_2$ be the eigenvalues of $A$. Then, the eigenvalues of $A-A^2$ are $\lambda_1-\lambda_1^2$ and $\lambda_2-\lambda_2^2$.
The eigenvalues of $\begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix}$ are $\dfrac{7 \pm \sqrt{5}}{2}$. Hence $\lambda_1-\lambda_1^2 = \dfrac{7 - \sqrt{5}}{2}$ and $\lambda_2-\lambda_2^2 = \dfrac{7 + \sqrt{5}}{2}$.
Solving gives us two possibilities for each eigenvalue:
$\lambda_1 = \dfrac{1}{2} \pm i\dfrac{1}{2}\sqrt{13-2\sqrt{5}}$ and $\lambda_2 = \dfrac{1}{2} \pm i\dfrac{1}{2}\sqrt{13+2\sqrt{5}}$.
If the entries of $A$ are all real, then either $\lambda_1, \lambda_2 \in \mathbb{R}$ or $\lambda_1 = \overline{\lambda_2}$, which is not the case here.
Hence, there is no real $2 \times 2$ matrix $A$ such that $A-A^2 = \begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Prove this identity... $$\frac{\sin 2x}{1+\cos 2x} \times \frac{\cos x}{1+\cos x}=\tan\frac{x}{2}$$
This is what I've done:
$$\frac{2\sin x \cos x}{1+\cos^2 x-\sin^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{2\sin x \cos x}{2\cos^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{\sin x}{1+\cos x}$$
I have no idea what to do next.
edit:
Solution:
$$\frac{\sin2\frac{x}{2}}{1+\cos2\frac{x}{2}}=$$
$$\frac{2\sin\frac{x}{2}cos\frac{x}{2}}{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}=$$
$$\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}=$$
$$\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2}$$
| $$\frac{\sin x}{1+\cos x}=\frac{\sin 2\frac{x}{2}}{1+\cos 2\frac{x}{2}}=\frac{2 \sin (\frac{x}{2})\cos(\frac{x}{2})}{1+\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find length of $CD$ where $\measuredangle BCA=120^\circ$ and $CD$ is the bisector of $\measuredangle BCA$ meeting $AB$ at $D$ $ABC$ is a triangle with $BC=a,CA=b$ and $\measuredangle BCA=120^\circ$. $CD$ is the bisector of $\measuredangle BCA$ meeting $AB$ at $D$. Then the length of $CD$ is ____ ?
A)$\frac{a+b}{4}$
B)$\frac{ab}{a+b}$
C)$\frac{a^2+b^2}{2(a+b)}$
D)$\frac{a^2+ab+b^2}{3(a+b)}$
I have tried using $AD+DB=AB$ and using cosine rule on each term. However it gets lengthy.
| Solution.
$\triangle ABC=\triangle ACD+\triangle BCD$
$\Longrightarrow$ $~\dfrac{ab\sin 120^{\circ}}{2}=\dfrac{b\cdot CD\sin 60^{\circ}}{2}+\dfrac{a\cdot CD\sin 60^{\circ}}{2}$
$\therefore$ $CD=\dfrac{ab}{a+b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/842649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Product in terms of $n$ of $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \cdots \cdot \frac{2n-1}{2n}$ What is the following product in terms of $n$?
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \cdots \cdot \frac{2n-1}{2n}$$
Thank you.
| Given
$$
(1/2)(3/4)(5/6)(7/8)\cdots([2n-1]/2n)
$$
Thus
$$
\frac{1}{2} \frac{3}{4} \frac{5}{6} \cdots \frac{2n-1}{2n} = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots [2n-1] \cdot 2n}{2^{2n} \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdots 2n)^2 } = \frac{[2n]!}{2^n (n!)^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/842911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Problem of quadratic equation If $\alpha$ be a root of $ 4x^2 +2x -1 = 0 $ , prove that the other root is $4\alpha^3 - 3\alpha$ . I have tried to do it but of no success.[$4\alpha^3 -2\alpha$ = $\dfrac {-1}{2}$ and $4\alpha^4 - 3\alpha^2$ = $\dfrac {-1}{4}$ ] .How to prove it?
| Write $4x^3-3x=(4x^2+2x-1)(x-1/2)-(x+1/2)$.
Plug $x=\alpha$ and get $4\alpha^3 - 3\alpha=-\alpha-1/2$, which is the other root because the sum of the roots is $-1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/846203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
When is the sum of two squares the sum of two cubes When does $a^2+b^2 = c^3 +d^3$ for all integer values $(a, b, c, d) \ge 0$. I believe this only happens when: $a^2 = c^3 = e^6$ and $b^2 = d^3 = f^6$. With the following exception:
*
*$1^3+2^3 = 3^2 + 0^2$
Would that statement be correct?
Is there a general formula for when this happens?
| *
*$a=x^3-3x^2y-3xy^2+y^3,b=x^3+3x^2y-3xy^2-y^3,c=d=x^2+y^2.$
*$a=3(x^3-3xy^2),b=3(3x^2y-y^3),c=x^2+y^2,d=2(x^2+y^2)$
$\cdots$
See this post.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/847773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Simple Differentiation Problem Involving Area Radius and Circumference A stone is dropped into a pool of water, and the area covered by the spreading ripple increases at a rate of $4 m^2 s^{-1} $.
Calculate the rate at which the circumference of the circle formed is increasing 3 seconds after the stone is dropped.
My method:
$$ \begin{align}
& \cfrac{dA}{dt}=4\\
& A = \pi \times r \times r \ \text{so} \ \cfrac{dA}{dr} = 2 \times \pi \times r \\
& C = 2 \times \pi \times r \ \text{so} \ \cfrac{dC}{dr} =2 \times \pi \\
& \cfrac{dA}{dt} = \cfrac{dA}{dr} \times \cfrac{dr}{dt} \\
& 4 = 2 \times \pi \times (3) \cfrac{ dr}{dt} \\
& \text{so} \ \cfrac{dr}{dt} = \cfrac{2}{(3 \times \pi)} \\
& \text{now} \ \cfrac{dC}{dt} = \cfrac{dC}{dr} \times \cfrac{dr}{dt} \\
& \text{so} \ \cfrac{dC}{dt} = 2 \times \pi \times \cfrac{2}{(3 \times \pi)}
\end{align} $$
I'm getting $1.3333...$
But the answer is $2.05m/s$ , I don't understand how to get it...
| Start from the equation $$\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt}$$ using $A(r) = \pi r^2$. We get $dA/dr = 2 \pi r$ and $$4 = 2 \pi r \frac{dr}{dt} \text{ or } r \frac{dr}{dt} = \frac{2}{\pi}.$$ Write $$r \, dr = \frac{2}{\pi} \, dt$$ and integrate, obtaining $$\int_0^r \rho \, d \rho = \frac{r^2}{2} = \int_0^t \frac{2}{\pi} \tau \, d\tau = \frac{2}{\pi} t.$$ Writing $r$ in terms of $t$ we have $$r = 2 \sqrt{\frac{t}{\pi}}.$$ Now differentiating we obtain $$\frac{dr}{dt} = \frac{1}{\sqrt{\pi t}}$$ and substituting into $$\frac{dC}{dt} = \frac{dC}{dr} \frac{dr}{dt}$$ we get $$\frac{dC}{dt} = 2 \pi \frac{1}{\sqrt{\pi t}} = 2 \sqrt{\frac{\pi}{t}}.$$ Taking $t=3$ it results $$\frac{dC}{dt} \approx 2,046 \, \frac{\text{m}}{\text{s}}.$$ EDIT: Perhaps a (non-explicit integrating) way to see it is notice the following: $$\frac{d}{dt} (r(t))^2 = 2 r \frac{dr}{dt} \text{ and } \frac{d}{dt} \left( \frac{2}{\pi} t \right) = \frac{2}{\pi}.$$ Therefore this implies that $$\frac{1}{2} \frac{d}{dt} (r(t))^2 = \frac{d}{dt} \left( \frac{2}{\pi} t \right).$$ Since $r(0) = 0$ you can assert that $$\frac{1}{2} r^2 = \frac{2}{\pi} t$$ and the solution goes as above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/849920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to prove $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty$ How to prove $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty.$$
I try to do like $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{n+m=N}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{m=1}^{N-1} \frac{1}{m^2+(N-m)^2}$$
$$\frac{1}{m^2+(N-m)^2}\leq \frac{2}{N^2}$$
but it doesn't work.
| $$
\begin{align}
\sum_{m=1}^\infty\sum_{n=1}^m\frac1{m^2+n^2}
&\ge\sum_{m=1}^\infty\sum_{n=1}^m\frac1{2m^2}\\
&=\frac12\sum_{m=1}^\infty\frac1m\\
&=+\infty.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Determinant involving recurrence Evaluate
$$\left| A \right| = \left| {\matrix{
{x + y} & {xy} & 0 & \cdots & \cdots & 0 \cr
1 & {x + y} & {xy} & \cdots & \cdots & 0 \cr
0 & 1 & {x + y} & \cdots & \cdots & 0 \cr
\cdots & \cdots & \cdots & \cdots & \cdots & \vdots \cr
0 & \cdots & 0 & 1 & {x + y} & {xy} \cr
0 & \cdots & 0 & 0 & 1 & {x + y} \cr
} } \right|$$
And show that $\det(A) = \frac{x^{n+1}-y^{n+1}}{x-y}$ if $x\ne y$ and $\det(A) = (n+1)x^n$ if $x=y$.
I actually was able to get this recurrence formula:
$$D_n = (x+y)\cdot D_{n-1} + xy\cdot D_{n-2}$$
I tried to prove it by induciton, but the algebric calculation didn't bring me to the desired result.
| Hint: compute $D_1 = x+y, D_2 = x^2 + xy + y^2$.
Then, prove that $D_n = x^n + x^{n-1}y + \dots + xy^{n-1} +y^n$ via induction.
Eventually, as $(x^n + x^{n-1}y + \dots + xy^{n-1} +y^n)(x-y) = x^{n+1} - y^{n+1}$, you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/852177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How do you simplify an algebraic expression? Please explain how to simplify an expression that is similar to this one
$\displaystyle\frac{a+3}{6}+\frac{a-4}{4}+\frac{a+2}{-3}$
| Depending on what "simplify" means (teachers and test writers should avoid this word, because it is often not specific enough to avoid ambiguity), my last expression below might be what is intended, in contrast to what Varum Iyer and mfl ended with. I've also carried out the steps a little differently (and in great detail) to give you an alternate way of carrying this out.
$$\frac{a+3}{6}\; + \; \frac{a-4}{4} \; + \; \frac{a+2}{-3}$$
$$= \;\; \frac{1}{6}(a+3)\; + \; \frac{1}{4}(a-4) \; - \; \frac{1}{3}(a+2)$$
$$= \;\; \frac{1}{6} \cdot a \; + \; \frac{1}{6} \cdot 3\; + \; \frac{1}{4} \cdot a \; - \; \frac{1}{4} \cdot 4 \; - \; \frac{1}{3} \cdot a \; - \; \frac{1}{3} \cdot 2$$
$$= \;\; \frac{1}{6} \cdot a \; + \; \frac{3}{6} \; + \; \frac{1}{4} \cdot a \; - \; \frac{4}{4} \; - \; \frac{1}{3} \cdot a \; - \; \frac{2}{3}$$
$$= \;\; \left(\frac{3}{6} \; - \; \frac{4}{4} \; - \; \frac{2}{3} \right) \;\; + \;\; \left (\frac{1}{6} \cdot a \; + \; \frac{1}{4} \cdot a \; - \; \frac{1}{3} \cdot a \right) $$
$$= \;\; \left(\frac{3}{6} \; - \; \frac{4}{4} \; - \; \frac{2}{3} \right) \;\; + \;\; \left (\frac{1}{6} \; + \; \frac{1}{4} \; - \; \frac{1}{3} \right)a $$
$$= \;\; \left(\frac{6}{12} \; - \; \frac{12}{12} \; - \; \frac{8}{12} \right) \;\; + \;\; \left (\frac{2}{12} \; + \; \frac{3}{12} \; - \; \frac{4}{12} \right)a $$
$$= \;\; -\frac{14}{12} \;\; + \;\; \frac{1}{12}a$$
$$= \;\; -\frac{7}{6} \;\; + \;\; \frac{1}{12}a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/852362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\prod_{i=1}^{n-1} \sin(i\pi/n) = 2^{1-n} n$ without complex functions. Note that $i$ here refers to indexing variable, not $\sqrt{-1}$.
$$\prod_{i=1}^{n-1} \sin\left(\frac{i \pi}{n}\right) = 2^{1-n} n$$
This formula was used here to give an 'elementary' proof of product of diagonals = N. Mathworld is the only place I can find the formula and that website cites a personal communication with 'T. Drane'
I have tried to prove it on my own but this is the only progress I've made.
Letting $n=2k+1$, I can show that
$$\prod_{i=1}^{2k} \sin\left(\frac{i \pi}{2k+1}\right) = \prod_{i=1}^{k} \cos^2\left(\frac{(2i-1) \pi}{4k+2}\right)$$
I used the sine product formula $\quad\sin(a)\sin(b)=\dfrac{\cos(a-b)-\cos(a+b)}{2}\quad$ to pair of the first/last terms, etc. Then used the cosine double-angle formula $\quad \cos(2a)+1 = 2\cos^2(a)\quad$ on the resulting product.
I don't know where to go on from on here and I was just investigating the special case of odd $n$. Any ideas on what to do next? Simpler approach that would also apply to $n=2k$?
PS: It would be preferred if the proof avoided complex functions in order to complete the answer to the diagonals question reference earlier.
| Thanks to Lahtonen's suggestion to use Chebyshev's polynomials, I was able to answer the question. I used this paper as a reference on the properties of the polynomials.
I'll prove $\frac{\sin((n+1)\theta)}{\sin(\theta)} = U_n(\cos(\theta))$ using induction (the paper used complex numbers to prove it).
$$
\frac{\sin(\theta)}{\sin(\theta)} = 1 = U_0(\cos(\theta)), \qquad
\frac{\sin(2\theta)}{\sin(\theta)} = 2\cos(\theta) = U_1(\cos(\theta)), \\ {~}\\ {~}\\
\begin{aligned}
\frac{\sin((n+1)\theta)}{\sin(\theta)}
&= 2\cos(\theta)\frac{\sin(n\theta)}{\sin(\theta)} - \frac{\sin((n-1)\theta}{\sin(\theta)} \\
&= 2\cos(\theta)U_{n-1}(\cos(\theta))-U_{n-2}(\cos(\theta)) \\
&= U_n(\cos(\theta))
\end{aligned}$$
Using Identities 7/9 in the linked paper, it followed that
$$\sin((2m+1)\theta) = (-1)^mT_{2m+1}(\sin(\theta))$$
Since $T_{2m+1}$ is an odd polynomial, I can factor it as such:
$$\sin((2m+1)\theta) = \sin(\theta)R_{2m}(\sin(\theta))$$
where $R_{2m}$ is an even polynomial of degree $2m$ with the following properties:
*
*unique roots are $u_k = \sin\left(\frac{k\pi}{2m+1}\right)$ for $k =\pm 1, \ldots, \pm m$.
*constant term is $2m+1$
*leading coefficient is $(-1)^m 2^{2m}$
Therefore, $$\prod_{k=1}^m u_k u_{-k} = (-1)^m \frac{2m+1}{2^{2m}}$$
By symmetry properties of the sine function, $$\prod_{k=1}^{2m} \sin\left(\frac{k\pi}{2m+1}\right) = \frac{2m+1}{2^{2m}} \qquad \square$$
This proves the identity for when $n$ is an odd integer. I'll introduce an abbreviation for simplicity's sake: let $F(n) = \prod_{k=1}^{n-1} \sin(k\pi/n)$. We have shown that $F(2m+1) = 2^{-2m}(2m+1)$ Also note that
$$F(2m+1) = \prod_{k=1}^{m} \sin^2\left(\frac{k\pi}{2m+1}\right) = \prod_{k=1}^{m} \cos^2\left(\frac{(2m+1-2k)\pi}{4m+2}\right)$$
This will be useful soon. Now, let $n=2m$.
$$F(2m) = \prod_{k=1}^{m-1} \sin^2\left(\frac{k\pi}{2m}\right) = \prod_{k=1}^{m-1} \cos^2\left(\frac{(m-k)\pi}{2m}\right) = \prod_{k=1}^{m-1} \cos^2\left(\frac{k\pi}{2m}\right)$$
I'll need to split this in two cases again: $n=4m$ and $n=4m+2$.
$$\begin{aligned}
F(4m+2) &= \prod_{k=1}^{2m} \cos^2\left(\frac{k\pi}{4m+2}\right) \\
&= \prod_{k=1}^m \left(\cos\left(\frac{k\pi}{4m+2}\right)\cos^2\left(\frac{(2m+1-k)\pi}{4m+2}\right) \right)^2 \\
&= \prod_{k=1}^m \frac{1}{4} \cos^2 \left(\frac{(2m+1-2k)\pi}{4m+2}\right) \\
&= \frac{1}{2^{2m}}F(2m+1) \\
&= \frac{1}{2^{2m}}\cdot\frac{2m+1}{2^{2m}} &\square
\end{aligned}$$
and finally, assume that $F(n)=\frac{n}{2^{n-1}}$ for all $n<4m$,
$$\begin{aligned}
F(4m) &= \prod_{k=1}^{2m-1} \cos^2\left(\frac{k\pi}{4m}\right) \\
&= \frac{1}{2}\prod_{k=1}^{m-1} \left(\cos\left(\frac{k\pi}{4m}\right)\cos\left(\frac{(2m-k)\pi}{4m}\right) \right)^2 \\
&= \frac{1}{2}\prod_{k=1}^{m-1} \frac{1}{4}\cos^2\left(\frac{(m-k)\pi}{2m}\right) \\
&= \frac{1}{2^{2m-1}} F(2m) \\
&= \frac{1}{2^{2m-1}}\cdot \frac{2m}{2^{2m-1}} & \blacksquare
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc.
Using that I get this
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$
From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering that $a+b\geq 2\sqrt{ab}$
But by using this, I get the following...
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}} \leq 3 \times \sqrt[3]{\dfrac{abc}{8abc}} = \dfrac{3}{2}$
Everything seems so perfect because I get the value $\dfrac{3}{2}$ as required, but this method isn't valid due to the change in direction! What is going on?
Is there a way of proving this inequality otherwise then?
| If we denote $s:=a+b+c$, we have to minimize
$$\frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}$$
for $a,b,c>0$.
Using new variables $x:=a/s$, $y:=b/s$, $z:=c/s$ we can see, that this is equivalent to minimizing
$$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}$$
for $x+y+z=1$.
If we notice that the function
$$f(x)=\frac{x}{1-x}=-1+\frac1{1-x}=-1-\frac1{x-1}$$
is convex on the interval $(0,1)$, we have
$$\frac{f(x)+f(y)+f(z)}3 \ge f\left(\frac{x+y+z}3\right),$$
i.e.
$$f(x)+f(y)+f(z) \ge 3f(1/3).$$
By computing $f(1/3)=1/2$ we see, that the last inequality is precisely
$$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z} \ge \frac32.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/855283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
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Evaluation of a trigonometric series $$
\mbox{Question: Evaluate}\quad
\tan^{2}\left(\pi \over 16\right) + \tan^{2}\left(2\pi \over 16\right) + \tan^{2}\left(3\pi \over 16\right) + \cdots + \tan^{2}\left(7\pi \over 16\right)
$$
What I did:
Well I know that $\tan^{2}\left(7\pi/16\right)$ is the same as
$\cot^{2}\left(\pi/16\right)$. Thus this will repeat for all values up to
$\tan^{2}\left(4\pi/16\right)$.
However, I don't understand where to proceed from there.
| Like Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$,
$$\tan16\theta=\frac{\binom{16}1t-\binom{16}3t^3+\cdots+\binom{16}{13}t^{13}-\binom{16}{15}t^{15}}{\cdots}\text{ where }t=\tan\theta$$
If $\displaystyle\tan16\theta=0,16\theta=r\pi$ where $r$ is any integer
$\displaystyle\implies\theta=\frac{r\pi}{16}$ where $0\le r\le15$
So, the roots of $\displaystyle\binom{16}1t-\binom{16}3t^3+\cdots+\binom{16}{13}t^{13}-\binom{16}{15}t^{15}=0$ are $\displaystyle\tan r\theta$ where $\displaystyle0\le r\le15,r\ne8$ as $\displaystyle\tan\frac{8\cdot\pi}{16}$ is not finitely defined
So, the roots of $\displaystyle\binom{16}1-\binom{16}3t^2+\cdots+\binom{16}{13}t^{12}-\binom{16}{15}t^{14}=0$
or of $\displaystyle\binom{16}{15}t^{14}-\binom{16}{13}t^{12}+\cdots+\binom{16}3t^2-\binom{16}1=0$
are $\displaystyle\tan r\theta$ where $\displaystyle1\le r\le15,r\ne8$
But as $\displaystyle\tan\frac{(16-n)\pi}{16}=\tan\left(\pi-\frac{n\pi}{16}\right)=-\tan\frac{n\pi}{16},\tan^2\frac{(16-n)\pi}{16}=\tan^2\frac{n\pi}{16};$
the roots of $\displaystyle\binom{16}{15}u^7-\binom{16}{13}u^6+\cdots+\binom{16}3u-\binom{16}1=0$ are $\displaystyle \tan^2\frac{(16-n)\pi}{16}=\tan^2\frac{n\pi}{16}$ where $1\le n\le 7$ or $9\le r\le15$
Now Vieta's formula is inviting
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find maximum of $P$ Let $$P = \frac{{{x^2}}}{{{x^2} + yz + x + 1}} + \frac{{y + z}}{{x + y + z + 1}} - \frac{{1 + yz}}{9}.$$
Find maximum of $P$ where $x, y,z$ are nonnegative real numbers such that ${x^2} + {y^2} + {z^2} = 2$.
I guess $\max P=\dfrac{5}{9}$ when $x=y=1, z=0$. But I can not prove.
| Step 1. Prove that
$$
x^2+yz+x+1\geq x(x+y+z+1)
$$
by using the expansion of $(x-y-z)^2\geq 0$ and $x^2+y^2+z^2=2$.
Step 2. Observe that
$$
x+y+z\leq\sqrt{2(x^2+(y+z)^2)}=2\sqrt{1+yz}
$$
Step 3. By Step 1, we have
$$
P\leq\frac{x^2}{x(x+y+z+1)}+\frac{y+z}{x+y+z+1}-\frac{1+yz}{9}=1-Q
$$
with
$$
Q=\frac{1}{x+y+z+1}+\frac{1+yz}{9}.
$$
By step 2 we have
$$
Q\geq \frac{1}{2\sqrt{1+yz}+1}+\frac{1+yz}{9}.
$$
Let $t=\sqrt{1+yz}\geq 1$ and
$$
f(t)=\frac{1}{2t+1}+\frac{t^2}{9}
$$
We can find
$$
\min_{t\geq 0}f(t)=f(0)=\frac{4}{9}
$$
and so $\max P=\frac{5}{9}$ at $x=y=1, z=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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find an elementary matrix $E$ such that $EA=B$ Let matrix
$$A = \begin{pmatrix} 1 & 2 & 0 \\ -3 & 1 & 1 \\ 0 & 4 & 2 \end{pmatrix} $$
and
$$B = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 7 & 1 \\ 0 & 4 & 2 \end{pmatrix} $$
Find an elementary matrix $E$ such that $EA= B$.
I try with many
$$E = \begin{pmatrix} -1 & -2 & 0 \\ 1 & 2 & 0 \\ 3 & -1 & 1 \end{pmatrix} $$
but not correct
| An elementary matrix is one which differs from the identity matrix by one elementary row operation.
Note that $B$ is the matrix $A$ with three times the first row added to the second. So if we take the matrix
$$E=\begin{pmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$
and now consider
$$EA=\begin{pmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 2 & 0 \\ -3 & 1 & 1 \\ 0 & 4 & 2\end{pmatrix}=\begin{pmatrix}1 & 2 & 0 \\ 0 & 7 & 1 \\ 0 & 4 & 2\end{pmatrix}=B$$
$E$ was the identity matrix with three times the first row added to the second. This is easy to figure out because the matrix $B$ is linear combinations of the rows of $A$ and linear combinations of the columns of $E$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/858468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Going from closed form to recurrence relation If I had a closed form for a sequence that I suspect to represents a recurrence relation how would I determine the recurrence relation?
In particular, I have the sequence
$$a_n = \frac{1}{4}\bigg(\frac{3}{2}+\sqrt{2}\bigg)^n + \frac{1}{4}\bigg(\frac{3}{2}-\sqrt{2}\bigg)^n$$
This reminds me of the closed form for the Fibonacci sequence, so I would assume that the sequence would satisfy a recurrence relation of the form $a_n = ba_{n-1}+ca_{n-2}$ for some integers (or at least rationals) $b,c$. What is a method I could use to determine the recurrence relation or determine that no recurrence relation exists?
| Suppose we want to find a recurrence relation such that
$a_n = \left(\frac{3}{2}+\sqrt{2}\right)^n$ and
$a_n = \left(\frac{3}{2}-\sqrt{2}\right)^n$ are both solutions, which has the form
$$
a_n -b a_{n-1} -c a_{n-2} = 0
$$
If we assume that $a_n = r^n$ for some $r$ is a solution, we'd have
$$
r^2(a_{n-2}) - br(a_{n-2}) + ca_{n-2} = 0 \implies\\
r^2 - br - c = 0
$$
So, it suffices to find $b,c$ such that $r = \frac{3}{2} \pm \sqrt{2}$ are the roots of the equation $r^2 - br - c = 0$. The unique choice that works is
$$
b = 3\\
c = -1/4
$$
So, your sequence will be the unique solution to the recurrence problem
$$
a_n = 3a_{n-1} - \frac 14 a_{n-2}\\
a_0 = \frac 12, \quad
a_1 = \frac 34
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/863431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding domain of $\sqrt{4-2\sqrt{x^2 - 1}}$ Find the (maximum) domain of: $\sqrt{4-2\sqrt{x^2 - 1}}$
Well, I guess it is $2 - \sqrt2 \sqrt{x^2 - 1}$ = $2- \sqrt{2(x^2 - 1)}$
from here, $x^2 - 1 \ge 0$ and then $ x \ge 1$ and $ x \ge -1$
What do you guys think?
| $$x^2-1 \geq 0 \Rightarrow x^2 \geq 1 \Rightarrow x \geq 1 \text{ or } x \leq -1$$
Also:
$$4-2 \sqrt{x^2-1} \geq 0 \Rightarrow 4 \geq 2 \sqrt{x^2-1} \Rightarrow 2 \geq \sqrt{x^2-1} \Rightarrow 4 \geq x^2-1 \Rightarrow 5 \geq x^2 \\ \Rightarrow \sqrt{-5} \leq x \leq \sqrt{5}$$
So:
$$1 \leq x \leq \sqrt{5} \ \text{ or } -\sqrt{5} \leq x \leq -1$$
Therefore the domain is:
$$\{ x \in \mathbb{R}: -\sqrt{5} \leq x \leq -1 \text{ or } 1 \leq x \leq \sqrt{5} \}$$
| {
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"source": "stackexchange",
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How I could evaluate this :$ \sum_{n=1}^{\infty}({-1})^{n+1}n(tan^{-1}s-s+\frac{s^3}{3}+....({-1})^{n+1} \frac{s^{2n+1}}{2n+1}) $? let $s$ be a complex variable which $Re(s)>0$. Evaluate : $ \sum_{n=1}^{\infty}({-1})^{n+1}n(tan^{-1}s-s+\frac{s^3}{3}+....({-1})^{n+1}
\frac{s^{2n+1}}{2n+1}) $
I would be interest for any replies or any comments
| For $n \in \mathbb{Z}_{+}$, let $a_n$ be the expression
$\displaystyle\;\tan^{-1}s-s+\frac{s^3}{3} + \cdots + (-1)^{n+1}$. It has an integral representation:
$$\begin{align}
a_n &= \int_0^s \left[\frac{1}{1+t^2} - \left( 1 + ( -t^2) + \cdots + (-t^2)^n\right)\right]dt\\
&= \int_0^s \left[\frac{1}{1+t^2} - \frac{1-(-t^2)^{n+1}}{1+t^2}\right]dt\\
&= \int_0^s \frac{(-t^2)^{n+1}}{1+t^2} dt
\end{align}
$$
As a result, we can rewrite the series at hand as
$$\sum_{n=1}^\infty (-1)^{n+1} n a_n
= \sum_{n=1}^\infty \int_0^s \frac{n t^{2(n+1)}}{1+t^2} dt
= \sum_{n=1}^\infty \int_0^1 \frac{n s (st)^{2(n+1)}}{1+(st)^2} dt
\tag{*1}
$$
Notice for $|s| < 1$, the sequences of functions
$\displaystyle\;\sum_{n=1}^N \frac{n s (st)^{2(n+1)}}{1+(st)^2}\;$
converges uniformly and absolutely to
$\displaystyle\;\frac{s(st)^4}{(1+(st)^2)(1-(st)^2)^2}\;$
as $N \to \infty$ for $t$ over $[0,1]$, we can switch the order of summation and taking limit in $(*1)$ and
get
$$\begin{align}
\sum_{n=1}^\infty (-1)^{n+1} n a_n
&= \int_0^1 \frac{(st)^4}{(1+(st)^2)(1-(st)^2)^2} sdt
= \int_0^s \frac{t^4}{(1+t^2)(1-t^2)^2} dt\\
&= \frac14 \left[ \log\frac{1-s}{1+s} + \tan^{-1} s + \frac{s}{(1-s^2)}\right]\
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I solve $y' = \sin(x - y)$? How can I solve the differential equation:
$$y'=\sin(x-y)$$
Could I do this?
$$\frac{dy}{dx}= \sin x \cos y - \sin y \cos x$$
But, how would I continue?
| Let $z = x - y$. Thus, $\dfrac{dy}{dx} = 1 - \dfrac{dz}{dx}$ and
$$
\dfrac{dy}{dx} = \sin(x - y) \quad \Rightarrow \quad \dfrac{dz}{dx} = 1 - \sin z \quad \Rightarrow \quad \int dx = \int \dfrac{dz}{1 -\sin z}
$$
The next step is to change $u = \tan(z/2)$ so that $dz = \dfrac{2du}{1 + u^2}$. Note that
$$
\sin z = \dfrac{2\sin(z/2)\cos(z/2)}{\cos^2(z/2) + \sin^2(z/2)} = \dfrac{2\tan(z/2)}{1 + \tan^2(z/2)} = \dfrac{2u}{1 + u^2}
$$
Thus,
$$
x = \int \dfrac{\dfrac{2du}{1 + u^2}}{1 -\dfrac{2u}{1 + u^2}} =2\int \dfrac{du}{1 + u^2 - 2u} = 2\int \dfrac{du}{(1-u)^2} = -\dfrac{2}{1-u} + C \quad \Rightarrow
$$
$$
x = \dfrac{2}{\tan(z/2) - 1} + C = \dfrac{2}{\tan\biggl(\dfrac{x - y}{2}\biggr) - 1} + C
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$. Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$.
My attempt:
$p\mid a^2+ab+b^2 \implies p\mid (a-b)(a^2+ab+b^2)\implies p\mid a^3-b^3$
So, we have, $a^{3k}\equiv b^{3k}\mod p$ and by Fermat's Theorem we have, $a^{3k+1}\equiv b^{3k+1}\mod p$ as $p$ is of the form $p=3k+2$.
I do not know what to do next. Please help. Thank you.
| By lil' Fermat, $a^{3k+1}\cong b^{3k+1}\bmod p\implies a\cdot
b^{3k}\cong b\cdot b^{3k}\bmod p$. Thus $\begin{align} a\cong b\bmod p\implies a=b+pk \implies a^2+ab+b^2=(b+pk)^2+(b+pk)b+b^2=3b^2+3bpk+p^2k^2\end{align}$. But this expression is divisible by $p$. So $3b^2=-3bpk-p^2k^2+pl\implies p|3b^2\implies p|b\implies p|a$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Equation $\frac {x^2-x-6}{x^2+x -2} - \frac {2x-4}{x-1}$ I am trying to answer the problem as to how to show $\dfrac {x^2-x-6}{x^2+x -2} - \dfrac {2x-4}{x-1} = -1$
I have the key with the answer; the left-hand side is supposed to reduce to $ -1$ and have confirmed this is correct but which steps do I take to reach that answer?
| Note that $$\dfrac{x^2 - x - 6}{x^2 + x-2} = \frac{(x-3)(x+2)}{(x+2)(x - 1)} = \frac {x-3}{x-1}$$
Now, we have a common denominator: $$\frac{(x-3)}{(x-1)} - \frac{2x-4}{(x-1)}=$$ $$= \frac{(x-3) - (2x-4)}{(x-1)} = $$ $$ = \dfrac{-x+1}{x-1} = \dfrac{-(x-1)}{x-1} = -1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric Identities help How do you solve this? I can't figure out what I should do.
$$\sin ^4\left(A\right)+\cos ^2\left(A\right)=\cos ^4\left(A\right)+\sin ^2\left(A\right)$$
Also, why is this equal zero? Can someone explain how that simplifies to be zero?
$$\frac{\left(\frac{1}{\cos \left(x\right)}\right)-1}{\left(\frac{1}{\cos \left(x\right)}\right)+1}+\frac{\cos \left(x\right)-1}{\cos \left(x\right)+1}$$
| For the 1st problem there's a little trick: add zero.
\begin{array}{lll}\sin^4A+\cos^2A&=&\sin^4A+(-\cos^4A+\cos^4A)+\cos^2A\\
&=&(\sin^4A-\cos^4A)+\cos^4A+\cos^2A\\
&=&(\sin^2A-\cos^2A)(\sin^2A+\cos^2A)+\cos^4A+\cos^2A\\
&=&(\sin^2A-\cos^2A)+\cos^4A+\cos^2A\\
&=&\cos^4A+\sin^2A
\end{array}
For the other problem, the trick is to multiply by one.
\begin{array}{lll}
\frac{\left(\frac{1}{\cos \left(x\right)}\right)-1}{\left(\frac{1}{\cos \left(x\right)}\right)+1}+\frac{\cos \left(x\right)-1}{\cos \left(x\right)+1}&=&\frac{\left(\frac{1}{\cos \left(x\right)}\right)-1}{\left(\frac{1}{\cos \left(x\right)}\right)+1}\cdot\frac{\cos(x)}{\cos(x)}+\frac{\cos \left(x\right)-1}{\cos \left(x\right)+1}\\
&=&\frac{1-\cos(x)}{1+\cos(x)}+\frac{\cos \left(x\right)-1}{\cos \left(x\right)+1}\\
&=&\frac{1-\cos(x)}{1+\cos(x)}-\frac{1-\cos(x)}{1+\cos(x)}\\
&=&0
\end{array}
| {
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How prove this $3^{\frac{5^{2^n}-1}{2^{n+2}}}\equiv (-5)^{\frac{3^{2^n}-1}{2^{n+2}}}\pmod {2^{n+4}}$ Question:
show that:
$$3^{\frac{5^{2^n}-1}{2^{n+2}}}\equiv (-5)^{\frac{3^{2^n}-1}{2^{n+2}}}\pmod {2^{n+4}},n\geq 1$$
My idea: since
I have prove
$$5^{2^n}-1\equiv 0\pmod {2^{n+2}}$$
$$3^{2^n}-1\equiv 0\pmod {2^{n+2}}$$
so let
$$\frac{5^{2^n}-1}{2^{n+2}}=k_{1},\frac{3^{2^n}-1}{2^{n+2}}=k_{2},k_{1},k_{2}\in N$$
so we must prove
$$3^{k_{1}}-(-5)^{k_{2}}\equiv 0\pmod {2^{n+4}}$$
and How prove it?
Thank you
| We can check with an easy induction that the orders of $3$ and $-5$ modulo $2^{n+4}$ are $2^{n+2}$, so in order to check the equality it is enough to know $\frac {a^{2^n}-1}{2^{n+2}}$ modulo $2^{n+2}$, hence $a^{2^n}-1$ modulo $2^{2n+4}$ (where $a=3$ and $5$)
Suppose $a$ is odd and define for $n\ge 1$ the sequence $(u_n)$ by $a^{2^n} = 1 + u_n.2^{n+2} \pmod {2^{2n+4}}$. Call $v_n.2^{2n+4}$ the error term in the modulo.
Then $a^{2^{n+1}} = (a^{2^n})^2 = 1 + u_n.2^{n+3} + v_n.2^{2n+5} + u_n^2.2^{2n+4} \pmod {2^{2n+6}}$, so $u_{n+1} = u_n + (v_n \pmod 2).2^{n+2} + (u_n^2 \pmod 4).2^{n+1}$
This tells you that when $u_1$ is odd (this is equivalent to $a \equiv 3,5 \pmod 8$), to get $u_{n+1}$ you have to flip the last bit of $u_n$ and add a "random" bit of weight $2^{n+2}$.
So the sequence $u_n$ converges to a $2$-adic integer $u(a)$, and $u_n$ is $u(a) \pmod {2^{n+2}}$ with its last bit flipped.
You can even drop the modulo and deduce that the sequence $\frac 14 (a^{2^n}-1)/2^n$ converges $2$-adically to $u(a)$
Remember that for any integers $x$ and $y$, $(1+x)^y = 1 + yx + \frac{y(y-1)}2 x^2 + \frac{y(y-1)(y-2)}6 x^3 + \ldots $
Using an even $x$ and letting $y=2^n$,
the sequence $((1+x)^y - 1)/y = x + \frac{y-1}2x^2+\frac{(y-1)(y-2)}6x^3 + \ldots$
converges $2$-adically to $x - x^2/2 + x^3/3 - x^4/4 + x^5/5 - \ldots$
which we will call $\log(1+x)$
Going back to our $u$ and putting $x = a-1$ (which is even), we deduce that $u(a) = \frac 1 4 \log (a)$.
So far we have determined that for $a \equiv 3,5 \pmod 8$, $\frac{a^{2^n}-1}{2^{n+2}} \pmod {2^{n+2}} = \frac 14 \log(a) + 2^{n+1} \pmod {2^{n+2}}$.
Since $3$ and $-5$ are of order $2^{n+2}$ modulo $2^{n+4}$, their $2^{n+1}$th power has to be $-1$, so we can simplify your question to :
Is $3^{\frac 14 \log 5} \equiv (-5)^{\frac 14 \log 3} \pmod {2^{n+4}}$ ?
Actually you can define $x^y$ by the formula above in the $2$-adics as long as $x$ is odd, and you can define $\exp(x) = \sum \frac {x^n}{n!}$ when $x$ is a multiple of $4$.
They satisfy the useful relations $\exp(\log(x)) = x$ when $x \equiv 1 \pmod 4$, $(x^z)(y^z) = (xy)^z$, $\exp(x)^y = \exp(xy)$, $\log(xy) = \log(x)+\log(y)$.
In particular, $\log 1 = 0$, then $\log (-1) = 0$, and so $\log x = \log (-x)$, and $\exp(\log x) = -x$ when $x \equiv 3 \pmod 4$.
Finally, $(-1)^x = (-1)^{x \pmod 2}$
Then $3^{\frac 14 \log 5} = (-1)^{\frac 14 \log 5} \exp(\frac 14 \log 5 \log (-3)) = - \exp(\frac 1 4 \log 5 \log 3)$
And $(-5)^{\frac 14 \log 3} = (-1)^{\frac 14 \log 3} \exp (\frac 14 \log 5 \log 3) = - \exp(\frac 1 4 \log 5 \log 3)$
| {
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Proving the following trigonometric proportion $$\frac{a\sin(B-C)}{b^2-c^2}=\frac{b\sin(C-A)}{c^2-a^2}=\frac{c\sin(A-B)}{a^2-b^2}$$
A, B, C are angles of a triangle and a, b, c are the sides of a triangle.
I tried using various things such as sine rule and then replacing the various rations in terms of sides of triangle, however it produced no result. Urgent help is needed!
| By the Law of Sines, we have
$$a = d \sin A \qquad b = d \sin B \qquad c = d \sin C$$
where $d$ is the diameter of the triangle's circumcircle. As a result, the relation to prove becomes (after eliminating some $d$s)
$$\frac{\sin A \;\sin(B-C)}{\sin^2B-\sin^2C} = \frac{\sin B \;\sin(C-A)}{\sin^2C-\sin^2A} = \frac{\sin C \;\sin(A-B)}{\sin^2A-\sin^2B} \qquad (\star)$$
Consider $\sin A\;\sin(B-C)$, where we know $A = \pi - (B+C)$ so that $\sin A = \sin(B+C)$. By a Product-to-Sum Identity,
$$\begin{align}
\sin(B+C) \sin(B-C) &= \frac{1}{2}\left(\;\cos((B+C)-(B-C))- \cos((B+C)+(B-C)) \;\right) \\[4pt]
&= \frac{1}{2}\left( \cos 2C - \cos 2B \right) \\[4pt]
&= \frac{1}{2}\left( \; ( 1 - 2 \sin^2 C ) - ( 1 - 2 \sin^2 B )\; \right) \\[4pt]
&= \sin^2 B - \sin^2 C
\end{align}$$
Thus, the left-most expression of $(\star)$ reduces to $1$; by symmetry, the other sides do, as well, so that all three are in fact equal. (Note that the expressions in the original problem reduce to $1/d$.)
| {
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Find $\tan x $ if $\sin x+\cos x=\frac12$ It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $.
I tried all identities I know but I have no idea how to proceed. Any help would be appreciated.
| Hint :
\begin{align}
\sin x+\cos x&=\frac12\\
\sin x&=\frac12-\cos x\\
\sin^2 x&=\left(\frac12-\cos x\right)^2\\
1-\cos^2 x&=\frac14-\cos x+\cos^2 x\\
2\cos^2 x-\cos x-\frac34&=0\\
8\cos^2 x-4\cos x-3&=0
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Polynomial $f(x)$ degree problem. Suppose the polynomial $f(x)$ is of degree $3$ and satisfies $f(3)=2$, $f(4)=4$, $f(5)=-3$, and $f(6)=8$. Determine the value of $f(0)$.
How would I solve this problem? It seems quite complicated...
| Let $g(x)=f(x+3)$. Then the difference table for $g$ is given by
$2\;\;\; 4\;\; -3\;\; 8$
$\;\; 2 \;\; -7\;\; 11$
$\;\;\;-9\;\; 18$
$\;\;\;\;\;\;27,$
so $f(0)=g(-3)=2+2\binom{-3}{1}-9\binom{-3}{2}+27\binom{-3}{3}=2+2(-3)-9(6)+27(-10)=-328.$
Alternatively, $f(x)=g(x-3)=2+2\binom{x-3}{1}-9\binom{x-3}{2}+27\binom{x-3}{3}=\frac{9}{2}x^3-\frac{117}{2}x^2+245x-328$ and $f(0)=-328$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/878130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
taylor series of $\ln(1+x)$? Compute the taylor series of $\ln(1+x)$
I've first computed derivatives (up to the 4th) of ln(1+x)
$f^{'}(x)$ = $\frac{1}{1+x}$
$f^{''}(x) = \frac{-1}{(1+x)^2}$
$f^{'''}(x) = \frac{2}{(1+x)^3}$
$f^{''''}(x) = \frac{-6}{(1+x)^4}$
Therefore the series:
$\ln(1+x) = f(a) + \frac{1}{1+a}\frac{x-a}{1!} - \frac{1}{(1+a)^2}\frac{(x-a)^2}{2!} + \frac{2}{(1+a)^3}\frac{(x-a)^3}{3!} - \frac{6}{(1+a)^4}\frac{(x-a)^4}{4!} + ...$
But this doesn't seem to be correct. Can anyone please explain why this doesn't work?
The supposed correct answers are: $$\ln(1+x) = \int \left(\frac{1}{1+x}\right)dx$$
$$\ln(1+x) = \sum_{k=0}^{\infty} \int (-x)^k dx$$
| Note that
$$\frac{1}{1+x}=\sum_{n \ge 0} (-1)^nx^n$$
Integrating both sides gives you
\begin{align}
\ln(1+x)
&=\sum_{n \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\\
&=x-\frac{x^2}{2}+\frac{x^3}{3}-...
\end{align}
Alternatively,
\begin{align}
&f^{(1)}(x)=(1+x)^{-1} &\implies \ f^{(1)}(0)=1\\
&f^{(2)}(x)=-(1+x)^{-2} &\implies f^{(2)}(0)=-1\\
&f^{(3)}(x)=2(1+x)^{-3} &\implies \ f^{(3)}(0)=2\\
&f^{(4)}(x)=-6(1+x)^{-4} &\implies \ f^{(4)}(0)=-6\\
\end{align}
We deduce that
\begin{align}
f^{(n)}(0)=(-1)^{n-1}(n-1)!
\end{align}
Hence,
\begin{align}
\ln(1+x)
&=\sum_{n \ge 1}\frac{f^{(n)}(0)}{n!}x^n\\
&=\sum_{n \ge 1}\frac{(-1)^{n-1}(n-1)!}{n!}x^n\\
&=\sum_{n \ge 1}\frac{(-1)^{n-1}}{n}x^n\\
&=\sum_{n \ge 0}\frac{(-1)^{n}}{n+1}x^{n+1}\\
&=x-\frac{x^2}{2}+\frac{x^3}{3}-...
\end{align}
Edit: To derive a closed for for the geometric series, let
\begin{align}
S&=1-x+x^2-x^3+...\\
xS&=x-x^2+x^3-x^4...\\
S+xS&=1\\
S&=\frac{1}{1+x}\\
\end{align}
To prove in the other direction, use the binomial theorem or simply compute the series about $0$ manually.
| {
"language": "en",
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"answer_count": 4,
"answer_id": 2
} |
How prove $x^3+y^3+z^3-3xyz\ge C|(x-y)(y-z)(z-x)|$
let $x,y,z\ge 0$,and such
$$x^3+y^3+z^3-3xyz\ge C|(x-y)(y-z)(z-x)|$$
Find the maximum of the $C$
witout loss of we assume that $$x+y+z=1$$
I think
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=(x+y+z)^3-3(xy+yz+xz)(x+y+z)=1-3(yz+xz+xy)$$
then $$(x-y)(y-z)(x-z)=?$$
so I can't，It is said $$C_{max}=\sqrt{9+6\sqrt{3}}$$
| This is the standard technique for this. This one is a bit tricky, but anyways. Let WLOG $x\le y\le z$ and let $y=a+x,z=b+x$. Now plug this into
$$(x+y+z)\frac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2)\ge C|(x-y)(y-z)(z-x)|$$
$$\iff(3x+a+b)(a^2-ab+b^2)\ge Cab(a-b)\quad \text{since } a\ge b$$
Now this is linear in $x$. Hence we may take $x=0$ to achieve the minimum of LHS. This gives
$$(a^3+b^3)\ge Cab(a-b)$$ We take $a=tb$ with $t\ge 1$.
Then $\displaystyle C_{\max}=\min_{t\ge 1}\dfrac{t^3+1}{t(t-1)}$
Which I believe is the one mentioned above. Is this working?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/878791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
The limit of $((1+x)^{1/x} - e+ ex/2)/x^2$ as $x\to 0$ $$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
*
*by directly substituting $x=0$ i got $\infty$
*by using L-H's rule, i got $-1/8$
the given options are
$a)\frac{24e}{11}$
$b)\frac{11e}{24}$
$c)\frac{e}{11}$
$d)\frac{e}{24}$
may be the question would be
$$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
| We have $$\begin{aligned}F(x)\,&= (1 + x)^{1/x} - e + \frac{ex}{2}\\
&= \exp\left(\frac{\log(1 + x)}{x}\right) - \exp\left(1 + \log\left(1 - \frac{x}{2}\right)\right)\\
&= \exp\left\{1 + \log\left(1 - \frac{x}{2}\right)\right\}\left\{\exp\left(\frac{\log(1 + x)}{x} - 1 - \log\left(1 - \frac{x}{2}\right)\right) - 1\right\}\\
&= e^{u}(e^{v} - 1)\end{aligned}$$ where $$u = 1 + \log\left(1 - \frac{x}{2}\right), v = \frac{\log(1 + x)}{x} - 1 - \log\left(1 - \frac{x}{2}\right)$$ and $u \to 1, v \to 0$ as $x \to 0$. Thus we have $$\begin{aligned}L \,&= \lim_{x \to 0}\frac{F(x)}{x^{2}}\\
&= \lim_{x \to 0}\frac{e^{u}(e^{v} - 1)}{v}\cdot\frac{v}{x^{2}}\\
&= e\cdot\lim_{x \to 0}\frac{1}{x^{2}}\left\{\frac{\log(1 + x)}{x} - 1 - \log\left(1 - \frac{x}{2}\right)\right\}\\
&= e\cdot\lim_{x \to 0}\frac{1}{x^{2}}\left\{\left(1 - \frac{x}{2} + \frac{x^{2}}{3} - \cdots\right) - 1 + \left(\frac{x}{2} + \frac{x^{2}}{8} + \frac{x^{3}}{24} + \cdots\right)\right\}\\
&= e\left(\frac{1}{3} + \frac{1}{8}\right) = \frac{11e}{24}\end{aligned}$$ We can also use LHR instead of series expansion for logs and this will require two applications of LHR to get the following expression $$\begin{aligned}L\, &= \frac{e}{6}\cdot\lim_{x \to 0}\frac{1}{x}\left(-\frac{1}{(1 + x)^{2}} + \frac{4 - x}{(2 - x)^{2}}\right)\\
&= \frac{e}{6}\cdot\lim_{x \to 0}\frac{1}{x}\frac{(4 - x)(1 + x)^{2} - (2 - x)^{2}}{(1 + x)^{2}(2 - x)^{2}}\\
&= \frac{e}{6}\cdot\lim_{x \to 0}\frac{1}{x}\frac{x(11 + x - x^{2})}{(1 + x)^{2}(2 - x)^{2}}\\
&= \frac{11e}{24}\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/879430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$
If $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$$
Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.
Things I have tried so far:
Using Cauchy inequality I can write:$$\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)(a+b+c) \geq \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2$$
but I can't continue this.I tried expanded form:$$\sum \limits_{cyc} \frac{a^5c^2}{a^2b^2c^2} \geq \sum \limits_{cyc} \frac{a^3c}{abc}$$
Which proceeds me to this Cauchy:$$\sum \limits_{cyc} \frac{a^5c^2}{abc}\sum \limits_{cyc}a(abc)\geq \left(\sum \limits_{cyc}a^3c\right)^2$$
I can't continue this one too.
The main Challenge is $3$ fraction on both sides which all of them have different denominator.and it seems like using Cauchy from first step won't leads to anything good.
| By AM-GM, we have:
$$\frac{a^3}{b^2} + a\ge \frac{2a^2}{b}\Rightarrow \frac{a^3}{b^2}\geq \frac{2a^2}{b} -a$$
Similar: $$\frac{b^3}{c^2}\geq \frac{2b^2}{c} -b$$
$$\frac{c^3}{a^2}\geq \frac{2c^2}{a} -c$$
Now adding them together and using C-S, we have:
$$\frac{a^3}{b^2}+\frac{b^3}{c^2} + \frac{c^3}{a^2}$$
$$\geq (\frac{a^2}{b} + \frac{b^2}{c}+\frac{c^2}{a})+(\frac{a^2}{b} + \frac{b^2}{c}+\frac{c^2}{a}) - (a+b+c)$$
$$\geq (\frac{a^2}{b} + \frac{b^2}{c}+\frac{c^2}{a})+\frac{(a+b+c)^2}{a+b+c} - (a+b+c)$$
$$= \frac{a^2}{b} + \frac{b^2}{c}+\frac{c^2}{a}$$
Equality holds when $a=b=c$
P.s: I don't think it's a nice solution:(
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What are the roots of $\sin(ax) + \sin((a + 2)x)$? I was playing around with $\sin(5x) + \sin(7x)$, wondering where the roots of the function are. I graphed it on wolframalpha and from the list of solutions I guessed that the solutions to $\sin(5x) + \sin(7x) = 0$ are exactly the roots of $\sin(6x)$.
After playing around with similar values, I have to wonder, is it true that solving
$$\sin(a\cdot x) + \sin((a + 2)\cdot x) = 0$$
is equivalent to
$$\sin((a + 1)\cdot x) = 0$$
This problem seems to want induction.
It is true that the roots of $\sin(x) + \sin(3x)$ are the same roots as $\sin(2x)$ (again checked via wolframalpha). So that takes care of the base case.
However, I have no idea how to get from the $\sin(a\cdot x) + \sin((a+2)\cdot x)$ to the $\sin((a+1)\cdot x)$ for any $a$. It's just been guesswork.
Is it true that $\sin(a\cdot x) + \sin((a + 2)\cdot x)$ has the same roots as $\sin((a + 1)\cdot x)$?
| We transform that sum into a product. An easy way to deduce it is: We begin with $\sin X + \sin Y$. Let $X = c + d$ and $Y = c - d$. This gives us $c = \frac{X + Y}{2}$ and $d = \frac{X - Y}{2}$. This way: $$\begin{align} \sin X + \sin Y &= \sin(c+d)+ \sin(c-d) \\ &= \sin c \cos d + \sin c \cos d + \sin c \cos d - \sin d \cos c \\ &= 2 \sin c \cos d \\ &= 2 \sin\left(\frac{X+Y}{2}\right)\cos\left(\frac{X-Y}{2}\right)\end{align}$$
Making $X = ax$ and $Y = (a+2)x$ gives: $$\sin(ax) + \sin((a+2)x) = 2\sin((a+1)x))\cos(-x) = 2\sin((a+1)x)~\cos(x)$$
since $\cos$ is an even function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/884295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluating limit $\lim_{x\to0_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}$ Find the limit $$\lim_{x\to0_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}.$$
First I've rewritten it as $$\sqrt{ \frac{2x(1-x)}{\sin x}} \frac{\log(1-x^2)}{1-\cos x}$$
Now we can use the approximation $\sin x = x + O(x^3)$ to see that the left factor will converge to $\sqrt{2}$, right? Now I've used L'Hospital and the same approximation for $\sin x$ to find that the right factor converges to $-2$. So the original limit is $-2 \sqrt{2}$.
Is this correct? Is there a better way to do this?
| You've found the right approach. I would use your knowlege about the Taylor approximations to intelligently split the limit up into 2 simpler limits then be more formal about solving it.
$$\begin{align}
\lim_{x\to0_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}} &= \lim_{x\to0_+} \left(\sqrt{ \frac{2x(1-x)}{\sin x}} \frac{\log(1-x^2)}{1-\cos x}\right)\\
&= \lim_{x\to0_+} \sqrt{ \frac{2x(1-x)}{\sin x}} \cdot \lim_{x\to0_+}\frac{\log(1-x^2)}{1-\cos x}\\
&= \sqrt{\lim_{x\to0_+} \frac{2x(1-x)}{\sin x}} \cdot \lim_{x\to0_+}\frac{\log(1-x^2)}{1-\cos x}\\
\text{L'Hopital on both limits}\\
&= \sqrt{\lim_{x\to0_+} \frac{2 - 4x}{\cos x}} \cdot \lim_{x\to0_+}\frac{\dfrac{-2x}{1 - x^2}}{\sin x}\\
&= \sqrt{2} \cdot \lim_{x\to0_+}\frac{-2x}{\sin x(1-x^2)}\\
\text{L'Hopital again}\\
&= \sqrt{2} \cdot \lim_{x\to0_+}\frac{-2}{\cos x - 2x\sin x - x^2 \cos x}\\
&= -2\sqrt{2}
\end{align}
$$
Now that I look at your question again I realize this is pretty much exactly how you explained your solution. Oh well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/884731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Evaluate $\sum_{k=0}^{511}\frac{\sin\frac\pi{2^{11}}}{\sin\frac{(4k+1)\pi}{2^{12}}\sin\frac{(4k+3)\pi}{2^{12}}}$ I need to evaluate
$$\sum_{n=0}^{511}\frac{\sin\frac\pi{2^{11}}}{\sin\frac{(4n+1)\pi}{2^{12}}\sin\frac{(4n+3)\pi}{2^{12}}}$$
Please give me some hint!
The final answer is $2^{10}$.
By CuriousGuest's answer, we need to prove
$$\cot\frac{\pi}{2^{12}}-\cot\frac{2046\pi}{2^{12}}+\cot\frac{2\pi}{2^{12}}-\cot\frac{2047\pi}{2^{12}}=2^{10}$$
any idea?
| Using the partial fractional decomposition, the sum can be restated as
$$\huge {\sum_{k=0}^{511} \cot{\frac{4k+1}{2^{12}}\pi} - \cot{\frac{4k+3}{2^{12}}\pi}}$$
Summing in reverse order gives
$$\huge {\sum_{k=0}^{511} \tan{\frac{4k+3}{2^{12}}\pi} - \tan{\frac{4k+1}{2^{12}}\pi}}$$
Using the identity $\cot{\theta} - \tan{\theta} \equiv 2\cot{2\theta}$
Gives us
$$\huge {\sum_{k=0}^{511} \cot{\frac{4k+1}{2^{11}}\pi} - \cot{\frac{4k+3}{2^{11}}\pi}}$$
Repeat this process until the denominator is equal to $4$.
The observant reader will realise that another application results in a vanishing sum, and therefore, is invalid.
$$\huge {\sum_{k=0}^{511} \cot{\left(k+\frac{1}{4}\right)\pi} - \cot{\left(k+\frac{3}{4}\right)\pi}}$$
$$\huge {\sum_{k=0}^{511} 1-(-1)} = 512 \times 2 = 1024$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How do I go from this $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$? So I am doing $\int\frac{x^2-3}{x^2+1}dx$ and on wolfram alpha it says the first step is to do "long division" and goes from $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$. That made the integral much easier, so how would I go about doing that in a clear manner? Thanks in advance for the help!
| Notice that:
$$
\frac{x^2 - 3}{x^2 + 1} = \frac{(x^2 + 1) - 4}{x^2 + 1}
= \frac{x^2 + 1}{x^2 + 1} - \frac{4}{x^2 + 1}
= 1 - \frac{4}{x^2 + 1}
$$
The hard part is the first step. Basically, we want the denominator to show up in the numerator somehow so that they cancel out nicely. So we apply wishful thinking and try to force the denominator to show up somehow by any means necessary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
How to prove this determinant is $\pi$?
prove or disprove
$$\pi=\begin{vmatrix}
3&1&0&0&0&\cdots\\
-1&6&1&0&0&\cdots\\
0&-1&\dfrac{6}{3^2}&1&0&\cdots\\
0&0&-1&\dfrac{3^2\cdot 6}{5^2}&1&\cdots\\
0&0&0&-1&\dfrac{5^2\cdot 6}{3^2\cdot 7^2}&\cdots\\
0&0&0&0&-1&\dfrac{3^2\cdot 7^2\cdot 6}{5^2\cdot 9^2}&\cdots\\
0&0&0&0&0&-1&\dfrac{5^2\cdot 9^2\cdot 6}{3^2\cdot 7^2\cdot 11^2}&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\ddots\\
\end{vmatrix}$$
I found maybe this is true.and is very interesting,(It seems Euler proved it?),because this follows from the Euler result:
$$\pi=3+\dfrac{1^2}{6+\dfrac{3^2}{6+\dfrac{5^2}{6+\dfrac{7^2}{6+\cdots}}}}$$
and can we solve it? Thank you
| A proof can be found on pages 11-13 of An Elegant Continued Fraction for $\pi$. Try deriving from the Leibnitz formula $1 - \frac{1}{3} + \frac{1}{5} - ... = \frac{\pi}{4}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
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Prove that $ \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $ Can someone please help me with this question?
$ \large \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $
My steps so far:
$ \large \frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{x-2}.4^{x}}{2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{2x-2}}{2^{x-2}} = 2^{3x-2}$
I get stuck here but I am assuming I need to get that 4 to a 2 somehow so I can combine them... so:
$ \large \frac {(2^2)^{2x-2}}{2^{x-2}} = 2^{3x-2}$
I feel like I am not on the right track here as I have no idea where to go now. Could anyone help please? Thank you!
| $\frac{4^{2x-2}}{2^{x-2}} = \frac{(2^2)^{2x-2}}{2^{x-2}} = 2^{(4x-4)-(x-2)} = 2^{3x-2}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Trigonometric Proof: Question:
If $m\cos\alpha-n\sin\alpha=p$ then prove that $m\sin\alpha+n\cos\alpha=\pm \sqrt{m^2+n^2-p^2}$
My Efforts:
$(m\cos\alpha-n\sin\alpha)^{2}=p^2$
$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$
Now i think we need to add something on both side but i can't figure out what to add
| $$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$$
multiply by $-1$ and add $m^2+n^2$ to both sides :
$$m^2+n^2-m^2\cos^2\alpha-n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=m^2+n^2-p^2$$
$$m^2(1-\cos^2\alpha)+n^2(1-\sin^2\alpha)-2mn\cos\alpha\ \sin\alpha=m^2+n^2-p^2$$
$$m^2(\sin^2\alpha)+n^2(\cos^2\alpha)-2mn\cos\alpha\ \sin\alpha=m^2+n^2-p^2$$
$$(m\sin\alpha+n\cos\alpha)^2=m^2+n^2-p^2$$
$$m\sin\alpha+n\cos\alpha=\pm \sqrt{m^2+n^2-p^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluation of limit How to evaluate the value of this limit?
$$\lim_{x\to 2} \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{x^2 - 4}}$$
Actually I'm struck at algebraic part. Please guide..
| $$\eqalign{ \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{x^2 - 4}} &= \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{(x+2)(x-2)}} \\&= \frac{1}{\sqrt{x+2}} + \frac{\sqrt x - \sqrt2}{\sqrt{(x+2)(x-2)}} \\&= \frac{1}{\sqrt{x+2}} + \frac{(\sqrt x - \sqrt2)(\sqrt x + \sqrt 2)}{\sqrt{(x+2)(x-2)}(\sqrt x + \sqrt 2)} \\&= \frac{1}{\sqrt{x+2}} + \frac{x - 2}{\sqrt{(x+2)(x-2)}(\sqrt x + \sqrt 2)} \\&= \frac{1}{\sqrt{x+2}}+\frac{\sqrt{x-2}}{\sqrt{x+2}(\sqrt x + \sqrt 2)},}$$
which is easy to find the limit of.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/888034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Is $\tan(x)$ differentiable for $x\in ( -\pi/2 , \pi/2 )$ This is an assignment question and in class we taught the definition that:
A function $f(x)$ is differentiable if we can find $f(x+h) - f(x) = Kh +h E(x,h)$, where $K=f'(x)$ and $E(x,h) \rightarrow 0$ as $h \rightarrow 0$
E.g. $f(x) = x^3 \Rightarrow f(x+h) - f(x) = (x+h)^3 - x^3 = 3x^2h+2xh^2 +h^3$
and so $3x^2h+2xh^2+h^3 =Kh +h E(x,h) $. Thus we can conclude that $K = 3x^2$ and $E(x,h) = 2xh + h^2 $, and thus $f$ is differentiable since $E(x,h) = 2xh + h^2 \rightarrow 0$ as $h \rightarrow 0$
But for proving that $tan(x)$ is differentiable I started and got stuck:
$\tan(x+h) - \tan(x) = \dfrac{\tan(h)(1+\tan^2(x))}{1-\tan(x)\tan(h)}$, I really want to find $K = 1+ \tan^2(x)$ but I can't find a way to get $(1+\tan^2(x))h$ without a fraction term, and $E(x,h) = 0 , h \rightarrow 0$.
Any hints or help would be great, thanks.
| We can get our hands dirty with more trig identities:
\begin{align*}
\frac{(\tan h)(1+\tan^2 x)}{1-\tan x\tan h}
&= \frac{\frac{\sin h}{\cos h}(\sec^2 x)}{1 - \frac{\sin x\sin h}{\cos x \cos h}} \cdot \frac{\cos x \cos h}{\cos x \cos h} \\
&= \frac{(\sin h)(\sec^2 x)(\frac{1}{\sec x})}{\cos x \cos h - \sin x\sin h} \\
&= \frac{\sin h \sec x}{\cos(x + h)} \\
&= \sin h \sec x \sec (x + h) \\
\end{align*}
Dividing this expression by $h$ and taking the limit as $h \to 0$, we obtain:
$$
\frac{d}{dx}\tan x = \left[\lim_{h \to 0} \frac{\sin h}{h} \right] \cdot \left[\lim_{h \to 0} \sec{x}\sec(x + h)\right] = 1 \cdot \sec x \sec(x + 0) = \sec^2x
$$
as desired.
| {
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The inflection points of $f(x)=(x^2-4x+1)e^{-x}$ I got the function $f(x)=(x^2-4x+1)e^{-x}$. The task is to find the inflection points. The correct answer is $x=4-\sqrt{5}$ and $ x=4+\sqrt{5} $. I got the second derivative to $f(x)$.
But when I equal it to zero I got $x=2+\sqrt{5}$ and $ x=2 - \sqrt{5} $.
What am I doing wrong?
| We have:
$$f(x) = P(x) e^x,$$
so $f'' = P'' e^x + 2 P' e^x + e^x P = e^x (2 + 4 x - 8 + x^2- 4x +1 ) = (x^2- 5) e^x$. Since $e^x > 0 $ for $x \in (-\infty,\infty)$, the solution of $f''(x) = 0$ is given by $x^2 -5 =0 $, so the inflection points are $x_i = \pm \sqrt{5}$.
Hope this helps.
Cheers!
Edit: have a look of the graph of $f(x)$ here.
| {
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Integral $\int_0^\infty \frac{x^n - 2x + 1}{x^{2n} - 1} \mathrm{d}x=0$ Inspired by some of the greats on this site, I've been trying to improve my residue skills. I've come across the integral
$$\int_0^\infty \frac{x^n - 2x + 1}{x^{2n} - 1} \mathrm{d}x=0$$
where $n$ is a positive integer that is at least $2$.
With non-complex methods, I know that the integral is $0$. But I know that it can be done with residue theorem.
The trouble comes in choosing a contour. We're probably going to do some pie-slice contour, perhaps small enough to avoid any of the $2n$th roots of unity, and it's clear that the outer-circle vanishes. But I'm having trouble getting the cancellation for the integral.
Can you help? (Also, do you have a book reference for collections of calculations of integrals with the residue theorem that might have similar examples?)
| Resolving the integrand into two partial fractions, we have
$$\displaystyle I=\int_0^{\infty} \frac{x^n-2 x+1}{\left(1+x^n\right)\left(1-x^n\right)} d x=\int_0^{\infty}\left(\frac{x}{1+x^n}-\frac{1-x}{1-x^n}\right) d x=J-K\tag*{} $$
For the integral $J$, we are going to transforms it into a Beta function by letting $y=\frac{1}{1+x^n}$.
$$\displaystyle \begin{aligned}J & =\frac{1}{n} \int_0^1 y^{-\frac{2}{n}}(1-y)^{\frac{2}{n}-1} d y \\& =\frac{1}{n} B\left(-\frac{2}{n}+1, \frac{2}{n}\right) \\& =\frac{\pi}{n} \csc \left(\frac{2 \pi}{n}\right) \quad \textrm{ (By Euler Reflection Formula)}\end{aligned}\tag*{} $$
Next, we are going to evaluate the integral
$ \displaystyle K=\displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{} $
by the theorem
$ \displaystyle \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \textrm{ where } z\notin Z.\tag*{} $
We first split the integral into two integrals
$ \displaystyle \displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{1-x}{1-x^{n}} d x+\int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{}$
Transforming the latter integral by the inverse substitution $ x\mapsto \frac{1}{x}$ m, we have
$\displaystyle \displaystyle \int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{x^{n-3}-x^{n-2}}{1-x^{n}} d x \tag*{} $
Putting back yields
$ \begin{aligned}\displaystyle K&=\int_{0}^{1} \frac{1-x+x^{n-3}-x^{n-2}}{1-x^{n}} d x\\\displaystyle &=\int_{0}^{1}\left[\left(1-x+x^{n-3}-x^{n-2}\right) \sum_{k=0}^{\infty} x^{n k}\right] d x\\ \displaystyle & =\sum_{k=0}^{\infty} \int_{0}^{1}\left[x^{n k}-x^{n k+1}+x^{n(k+1)-3}-x^{n(k+1)-2}\right] d x\\ & =\sum_{k=0}^{\infty}\left(\frac{1}{n k+1}-\frac{1}{n k+2}+\frac{1}{n(k+1)-2}-\frac{1}{n(k+1)-1}\right)\\ & =\sum_{k=0}^{\infty}\left[\frac{1}{n k+1}-\frac{1}{n(k+1)-1}\right]+\sum_{k=0}^{\infty}\left[\frac{1}{n(k+1)-2}-\frac{1}{n k+2}\right] \end{aligned}\tag*{} $
Modifying yields
$$\displaystyle \begin{aligned} K&=\frac{1}{n}\left[\sum_{k=0}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-1}^{-\infty} \frac{1}{k+\frac{1}{n}}\right]+\frac{1}{n}\left(\sum_{k=1}^{\infty} \frac{1}{k-\frac{2}{n}}+\sum_{k=0}^{-\infty} \frac{1}{k-\frac{2}{n}}\right)\\& =\frac{1}{n}\left(\sum_{k=-\infty}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-\infty}^{\infty} \frac{1}{k-\frac{2}{n}}\right)\end{aligned} \tag*{} $$
By the Theorem,
$$ \displaystyle \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \tag*{} $$
where $ \displaystyle z\notin Z,$
we have
$$ \displaystyle \displaystyle K=\frac{1}{n}\left[\pi \cot \left(\frac{\pi}{n}\right)+\pi \cot \left(\frac{-2 \pi}{n}\right)\right]=\frac{\pi}{n}\left[\cot \left(\frac{\pi}{n}\right)-\cot \left(\frac{2 \pi}{n}\right)\right] =\frac{\pi}{n} \csc \frac{2 \pi}{n} =J\tag*{} $$
We can now conclude that
$\displaystyle \boxed{I=J-K=0 }\tag*{} $
| {
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The other ways to calculate $\int_0^1\frac{\ln(1-x^2)}{x}dx$
Prove that
$$\int_0^1\frac{\ln(1-x^2)}{x}dx=-\frac{\pi^2}{12}$$
without using series expansion.
An easy way to calculate the above integral is using series expansion. Here is an example
\begin{align}
\int_0^1\frac{\ln(1-x^2)}{x}dx&=-\int_0^1\frac{1}{x}\sum_{n=0}^\infty\frac{x^{2n}}{n} dx\\
&=-\sum_{n=0}^\infty\frac{1}{n}\int_0^1x^{2n-1}dx\\
&=-\frac{1}{2}\sum_{n=0}^\infty\frac{1}{n^2}\\
&=-\frac{\pi^2}{12}
\end{align}
I am wondering, are there other ways to calculate the integral without using series expansion of its integrand? Any method is welcome. Thank you. (>‿◠)✌
| After substituting $y=x^2$, we obtain
$$
\int_0^1\frac{\ln(1-x^2)}{x}\ dx=\frac12\int_0^1\frac{\ln(1-y)}{y}\ dy
$$
Using the fact that
$$
\frac{\ln(1-x)}{x}=-\int_0^1\frac{1}{1-xy}\ dy
$$
then
$$
\frac12\int_0^1\frac{\ln(1-x)}{x}\ dx=-\frac12\int_{x=0}^1\int_{y=0}^1\frac{1}{1-xy}\ dy\ dx.
$$
Using transformation variable by setting $(u,v)=\left(\frac{x+y}{2},\frac{x-y}{2}\right)$ so that $(x,y)=(u-v,u+v)$ and its Jacobian is equal to $2$. Therefore
$$
-\frac12\int_{x=0}^1\int_{y=0}^1\frac{1}{1-xy}\ dy\ dx=-\iint_A\frac{du\ dv}{1-u^2+v^2},
$$
where $A$ is the square with vertices $(0,0),\left(\frac{1}{2},-\frac{1}{2}\right), (1,0),$ and $\left(\frac{1}{2},\frac{1}{2}\right)$. Exploiting the symmetry of the square, we obtain
$$
\begin{align}
\iint_A\frac{du\ dv}{1-u^2+v^2}=\ &2\int_{u=0}^{\Large\frac12}\int_{v=0}^u\frac{dv\ du}{1-u^2+v^2}+2\int_{u=\Large\frac12}^1\int_{v=0}^{1-u}\frac{dv\ du}{1-u^2+v^2}\\
=\ &2\int_{u=0}^{\Large\frac12}\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{u}{\sqrt{1-u^2}}\right)\ du\\
&+2\int_{u=\Large\frac12}^1\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)\ du.
\end{align}
$$
Since $\arctan\left(\frac{u}{\sqrt{1-u^2}}\right)=\arcsin u$, and if $\theta=\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)$ then $\tan^2\theta=\frac{1-u}{1+u}$ and $\sec^2\theta=\frac{2}{1+u}$. It follows that $u=2\cos^2\theta-1=\cos2\theta$ and $\theta=\frac12\arccos u=\frac\pi4-\frac12\arcsin u$. Thus
$$
\begin{align}
\iint_A\frac{du\ dv}{1-u^2+v^2}
&=2\int_{u=0}^{\Large\frac12}\frac{\arcsin u}{\sqrt{1-u^2}}\ du+2\int_{u=\Large\frac12}^1\frac{1}{\sqrt{1-u^2}}\left(\frac\pi4-\frac12\arcsin u\right)\ du\\
&=\bigg[(\arcsin u)^2\bigg]_{u=0}^{\Large\frac12}+\left[\frac\pi2\arcsin u-\frac12(\arcsin u)^2\right]_{u=\Large\frac12}^1\\
&=\frac{\pi^2}{36}+\frac{\pi^2}{4}-\frac{\pi^2}{8}-\frac{\pi^2}{12}+\frac{\pi^2}{72}\\
&=\frac{\pi^2}{12}
\end{align}
$$
and the result follows.
| {
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Dealing with absolute values after trigonometric substitution in $\int \frac{\sqrt{1+x^2}}{x} \text{ d}x$. I was doing this integral and wondered if the signum function would be a viable method for approaching such an integral. I can't seem to find any other way to help integrate the $|\sec \theta|$ term in the numerator of the integrand.
$$ \begin{aligned} \int \dfrac{\sqrt{x^2 + 1}}{x} \text{ d}x \ \ & \overset{x=\tan \theta}= \int \dfrac{\sqrt{\sec^2 \theta} \sec^2 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \int \dfrac{|\sec \theta| \sec^2 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \int \dfrac{\text{sgn} (\sec \theta) \sec^3 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \text{sgn} (\sqrt{1+x^2}) \left( - \log \left| \dfrac{\sqrt{1+x^2} + 1}{x} \right| + \sqrt{1+x^2} \right) + \mathcal{C} \end{aligned} $$
It's clear that $\text{sgn} (\sqrt{1+x^2}) = 1$ since the sign of the argument of that function is always positive and the signum function extracts the sign. So I'd leave the integral as: $$ \begin{aligned} \int \dfrac{\sqrt{x^2 + 1}}{x} \text{ d}x = \log \left| x \right| - \log \left( \sqrt{1+x^2} + 1 \right) + \sqrt{1+x^2} + \mathcal{C} \end{aligned} $$
Would that be ok?
Also, apparently Wolfram has suggested that my final result should have $x$ as opposed to $|x|$ in the argument of the first logarithm. Why is that?
Any help would be greatly appreciated!
| $ x = \sinh u$
$$S = \int \frac{\sqrt{1+x^2}}{x}\mathbb{d}x = \int \frac{\cosh^2 u}{\sinh u }\mathbb{d}u = \int \Big\{\mathbb{csch}\ u +\sinh u \Big\}\mathbb{d}u \\
=-\ln\Big(\mathbb{csch}\ u +\mathbb{coth}\ u\Big)+\cosh u + C\\
$$
$$\cosh \sinh^{-1} x = \frac{1}{2}\bigg\{x+\sqrt{x^2+1} + \frac{1}{x+\sqrt{x^2+1}}\bigg\} = \sqrt{1+x^2} $$
$$S= \ln x +\sqrt{1+x^2} -\ln\Big(1+\sqrt{x^2+1}\Big)+C $$
Clearly it's simply a convenience to have $|x|$.
| {
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How prove $(x+\sqrt{x^{2}-1})^{n}+(x-\sqrt{x^{2}-1})^{n}\leq 2(1+n(x-1))^{n}$ for $n\in\mathbb{N}$? Let $x\ge 1$. How prove that $(x+\sqrt{x^{2}-1})^{n}+(x-\sqrt{x^{2}-1})^{n}\leq 2(1+n(x-1))^{n}$ for $n\in\mathbb{N}$?
| The easy way:
Let $f(x)=RHS-LHS$.
$f'(x)=2n^2(1+n(x-1))^{n-1}-n(1+\frac{x}{\sqrt{x^2-1}})(x+\sqrt{x^2-1})^{n-1}-n(1-\frac{x}{\sqrt{x^2-1}})(x-\sqrt{x^2-1})^{n-1}$
$f'(x)=2n^2(1+n(x-1))^{n-1}-\frac{n}{\sqrt{x^2-1}}(x+\sqrt{x^2-1})^{n}+\frac{n}{\sqrt{x^2-1}}(x-\sqrt{x^2-1})^{n}$
$f'(x) > 2n^2(1+n(x-1))^{n-1}-\frac{n^2}{x+\sqrt{x^2-1}}(x+\sqrt{x^2-1})^{n}-\frac{n^2}{x-\sqrt{x^2-1}}(x-\sqrt{x^2-1})^{n}$.
That, by induction, gives $f'(x) \geq 0$, so we are done.
Verifying that $\frac{n}{\sqrt{x^2-1}}<\frac{n^2}{x+\sqrt{x^2-1}}$, for $n \geq 2, x>1$, and formalising the inductive argument, are exercises left to the reader.
| {
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Simplify the following compound fraction: $$\frac{2x+1}{\frac{3}{x^2}+\frac{2x+1}{x}}$$
My calculator says the final answer is $$\frac{x^2(2x+1)}{2x^2+x+3}$$
Please show the work. Thanks.
| The answer could be considered complete, but, as I said in my comments above, you could simplify a bit more since the numerator has a higher degree than the denominator, i.e.:$$
\frac{x^2(2x+1)}{2x^2+x+3}=\frac{2x^3+x^2}{2x^2+x+3}=\frac{2x^3+x^2+3x-3x}{2x^2+x+3}
$$$$=\frac{x(2x^2+x+3)-3x}{2x^2+x+3}=x-\frac{3x}{2x^2+x+3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Counting words with specific requirements How many words of length exactly 6 can be created from letters A,B,C,D,E,F, so that each word has not more than one A, not more than one B, not less than one C and not less than one D?
I've tried using simple binomial coefficient, but it is clear that using this method one would count the same word multiple times.
| We can break this up into 4 cases:
1) If the word has 0 A's and 0 B's, then there are $4^6$ possible words, and subtracting the words with no C's and no D's gives $4^6-2(3^6)+2^6=2702$ possibilities.
2) If the word has 1 A and 1 B, then there are 6 choices for the A, 5 choices for the B, and then $[4^4-2(3^4)+2^4]$ ways to complete the word so it contains at least one C and at least one D; so there are $6(5)[4^4-2(3^4)+2^4]=3300$ possibilities.
3) If the word contains 1 A and 0 B's, there are 6 choices for the A, and then
$[4^5-2(3^5)+2^5]$ ways to complete the word; so there are $6[4^5-2(3^5)+2^5]=3420$ possibilities.
4) If the word contains 1 B and 0 A's, then there are $6[4^5-2(3^5)+2^5]=3420$ possibilities as in case 3.
Thus there are $12,842$ possible words.
An alternate approach would be to use exponential generating functions:
Since $g_{e}(x)=(1+x)^{2}(e^{x}-1)^2e^{2x}=(1+2x+x^2)(e^{4x}-2e^{3x}+e^{2x})$,
the coefficient of $x^6$ in $g_{e}(x)$ is given by
$\displaystyle\frac{a_{6}}{6!}=\frac{4^6}{6!}-2\cdot\frac{3^6}{6!}+\frac{2^6}{6!}+2\cdot\frac{4^5}{5!}-4\cdot\frac{3^5}{5!}+2\cdot\frac{2^5}{5!}+\frac{4^4}{4!}-2\cdot\frac{3^4}{4!}+\frac{2^4}{4!}$;
so the number of possible words is given by
$a_{6}=4^6-2(3^6)+2^6+2(6)(4^5)-6(4)(3^5)+6(2)(2^5)+6(5)(4^4)-6(5)(2)(3^4)+6(5)(2^4)$.
| {
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Solve for positive reals Solve for positive reals $x,y$
$(x+y)(1+\frac{1}{xy})+4=2(\sqrt{2x+1}+\sqrt{2y+1})$
I started by accumulating the terms of $x$ and then used AM-GM inequality but unsuccessfully....
| First:
$$\begin{align}
(x+y)\left(1+\frac{1}{xy}\right)+4 &= x+y+\frac{1}y+\frac{1}x +4\\
&= \left(x+2+\frac{1}{x}\right)+\left(y+2+\frac{1}{y}\right) \\
&=\frac{1}{x} \left(x+1\right)^2 + \frac{1}{y}\left(y+1\right)^2
\end{align}$$
So, define $$f(x)=\frac{1}{x}\left(x+1\right)^2-2\sqrt{2x+1}$$
Then you want to find two positive values $x,y$ so that $f(x)=-f(y)$.
Can $f(u)$ ever be negative or zero?
Letting $g(x)=xf(x) = (x+1)^2-2x\sqrt{2x+1}$, you need to solve $g(x)\leq 0$.
So $(x+1)^4 \leq 4x^2(2x+1)$ or $$x^4-4x^3+2x^2+4x+1\leq 0$$
$x^4-4x^3+2x^2+4x+1$ factors (via Wolfram Alpha) as $(x^2-2x-1)^2$. So there are no negative values for $g$ and only one positive zero at $x=1+\sqrt 2$, and therefore a solution of $x=y=1+\sqrt 2$ to your original equation, and this is the only positive real solution.
| {
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How to prove that $\tan^2(\frac\theta2)= \tan^2(\frac\alpha2)\tan^2(\frac\beta2)$? I'm unable to solve this question:
$\cos(\theta)=\dfrac{\cos(\alpha)+\cos(\beta)}{1+\cos(\alpha) \cos(\beta)}$
Prove: $\tan^2\left(\frac\theta2\right)= \tan^2\left(\frac\alpha2\right)\tan^2\left(\frac\beta2\right)$
I have tried the following:
*
*Using the indentity: $\cos(\theta)=\dfrac{1-\tan^2\left(\frac\theta2\right)}{1+\tan^2\left(\frac\theta2\right)}$
*Diving by $\cos (\alpha) \cos (\beta)$
*Creating a triangle, to find $\tan(\theta)= \sin \alpha \sin \beta$
Every time I got a huge complex equation with roots. Any help would be appreciated.
| $$ \tan ^2 {\frac {\theta}{2}} = \frac{\sin^2 {\frac {\theta}{2}} }{\cos^2 {\frac {\theta}{2}}} = \frac{ \left({\dfrac{1 - \cos \theta}{2}}\right)}{\left({\dfrac{1 + \cos \theta}{2}}\right)} \;\; \text{(since $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta= 1 - 2 \sin^2 \theta = 2 \cos^2\theta - 1$ )}$$
Now substituting for $\cos \theta$ and factorising we have,
$$ = \frac{1 + \cos \alpha \cos \beta - \cos \alpha - \cos \beta }{ 1 + \cos \alpha \cos \beta + \cos \alpha + \cos \beta} = \frac{(1 - \cos \alpha)(1 - \cos \beta)}{(1 + \cos \alpha)(1 + \cos \beta)} $$
Now use the same formulas for $cos 2 \theta $ mentioned above strategically to negate the $1$ present in all the factors and to introduce $\dfrac {\alpha}{2} $. Then,
$$ = \frac{[1 - (1 - 2 \sin^2 \frac {\alpha}{2} ) ][1 - ( 1 - 2 \sin^2 \frac {\beta}{2})]}{[1 + (2 \cos^2 \frac {\alpha}{2} - 1)][1 + (2 \cos ^2 \frac {\beta}{2} - 1)]} $$
Remember $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$.
Now Simplify!!
| {
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Integrate $\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx$ Integrate: $$\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx$$
My attempt :
I substituted $\sec{x} + \tan{x} $ as $t^2$
Then, $$ (\sec{x} \cdot \tan{x} + \sec^2x) dx =2tdt$$
$$\sec{x}( \tan{x} + \sec{x}) dx =2tdt$$
$$\sec{x}\cdot t^2 dx =2tdt$$
But there is $\sec{x}$ left outside.
| If we use Weierstrass substitution, the problem becomes$$I=\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx=2 \int \frac{ \sqrt{\frac{1+t}{1-t}} \left(1+t^2\right)}{\left(1-t^2\right)^2}\,dt$$ Now, setting $$\sqrt{\frac{1+t}{1-t}}=z$$ the integral becomes $$I=\int (z^2+\frac{1}{z^2})\,dz=\frac{z^3}{3}-\frac{1}{z}$$ Now, back to $x$, we get $$I=\frac{2}{3} \Big(2 \sin (x)-1\Big) \sec (x) \sqrt{\frac{\sin \left(\frac{x}{2}\right)+\cos
\left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)-\sin
\left(\frac{x}{2}\right)}}$$
| {
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Evaluate $\int\frac{\sqrt{9-x^2}}{x^2}\mathrm dx$ I am trying to solve $$\int\frac{\sqrt{9-x^2}}{x^2}\mathrm dx$$
My answer is slightly different to the memo:
$x=3\sin\theta\quad\iff\quad\theta=\arcsin\left(\frac x 3\right)\\
\text dx=3\cos\theta\ \text d\theta\\$
$\begin{align}I&=\int\frac{3\sqrt{1-\sin^2\theta}}{3\sin^2\theta}\cdot3\cos\theta\ \text d\theta=3\int\frac{\cos^2\theta}{\sin^2\theta}\ \text d\theta=3\int\cot^2\theta\ \text d \theta\\
&=3\int\csc^2\theta\ \text d\theta - 3\int\text d \theta\\
&=-3\cot\theta-3\theta+C\\
&=-\frac{\sqrt{1-\left(\frac x 3\right)^2}}{\frac x 3}-3\arcsin\left(\frac x 3\right)+C\\
&=-\frac{3\sqrt{9-x^2}}{3x}-3\arcsin\left(\frac x 3\right)+C\\
&=-\frac{\sqrt{9-x^2}}x-3\arcsin\left(\frac x 3\right)+C \end{align}$
and the memo has $$-\frac{\sqrt{9-x^2}}x-\arcsin\left(\frac x 3\right)+C$$
Your help is appreciated!
| $$\int\frac{\sqrt{9-x^2}}{x^2}dx=\vert x=3\sin t\Rightarrow dx=-3\cos t dt\vert=\int\frac{\sqrt{9-9\sin^2t}}{9\sin^2 t}\cdot (-3\cos t) dt=$$$$\int\frac{\sqrt{9(1-\sin^2 t)}}{9\sin^2 t}\cdot (-3\cos t) dt=\int\frac{-3\sqrt{1-\sin^2 t}}{9\sin^2 t}\cdot 3\cos t dt=$$
$$-\int\frac{\cos^2 t}{\sin^2 t }dt=-\int\cot^2 t dt=-(-\cot t-t)=\cot t+t=$$
You need only to substitution $t$ for $t=\arcsin \frac{x}{3}$
| {
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Why is $ A_1 x + ... + A_n x^n $ a solution of $ \sum_0^{n} (-1)^n \frac{x^n}{n!} \frac{d^n y}{d x^n} = 0 $? I was playing(/fiddling) around with some maths and I saw this pattern(
where $ A_n $ is a constant.):
$ A_1 x $ is a soultion of:
$$ \frac{y}{x} - \frac{dy}{dx} = 0 $$
$ A_1 x + A_2 x^2 $ is a solution of:
$$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} =0 $$
$ A_1 x + A_2 x^2 + A_3 x^3 $ is a solution of:
$$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} - \frac{x^2}{3!} \frac{d^3y}{dx^3} =0 $$
It continues so on. Can someone prove the solution of
$ A_1 x + A_2 x^2 + A_3 x^3 + ... + A_n x^n $ is:
$$ \sum_{k=0}^{n} (-1)^k \frac{x^{k-1}}{k!} \frac{d^k y}{d x^k} = 0 $$
| Use induction. Suppose the property holds for
$$ A_1 x + A_2 x^2 + A_3 x^3 + ... + A_n x^n $$
Prove that it works for $ n+1$ terms by writing the expression, use the fact that $\dfrac{\mathrm d^{n+1}y}{\mathrm dx^{n+1}}f = \dfrac{\mathrm d^{n}y}{\mathrm dx^{n}}\left(\dfrac{\mathrm dy}{\mathrm dx}f\right)$ and then show that your new equation is equivalent to one with $n$ terms.
| {
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Find $n$ between $100$ and $1000$ so that $2^n+2$ is divisible by $n$ Find $n$ such that $n$ divides $2^n + 2$. Also, $n$ should be between $100$ and $1000$.
It can be easily seen that $n$ is not a multiple of $4$. By brute force I have figured out that answer is $946$, but I don't know how to proceed further.
| See OEIS sequence A006517 and references there.
EDIT: $n$ must be even, but not divisible by $4$, and it's not hard to show
that $n=2p^j$ for prime $p$ doesn't work, so $n$ is divisible by at least two odd primes.
Let's say we guess that $n = 2 p q$ for odd primes $p < q$. Since $n/2 < 500 < 23^2$, we must have $p < 23$. Thus we need $ 2^{2pq-1} + 1$ to be divisible by $p$ and by $q$. By Fermat's Little Theorem, $2^{2pq} \equiv 2^{2q} \mod p$, so
$p | 2^{2q-1}+1$ and $q | 2^{2p-1} + 1$. Try each possible $p$ in turn.
For $p = 3$, $2^{2p-1} + 1 = 33$ so $q = 11$: too small ($2pq = 66 < 100$).
For $p = 5$, $2^{2p-1} + 1 = 513 = 3^3 \times 19$ so $q = 19$,
but $2^{2q-1} \equiv (-1)^q \times 2^{-1} \equiv 2 \mod 5$, no good.
For $p = 7$, $2^{2p-1} + 1 = 8193 = 3 \times 2731$ so $q = 2731$, too big.
For $p = 11$, $2^{2p-1} + 1 = 2097153 = 3^2 \times 43 \times 5419$, so $q = 43$.
And $2^{2q-1} + 1 = 2^{85} + 1 \equiv 2^5 +1 \equiv 0 \mod 11$, so this one works: $n = 2 \times 11 \times 43 = 946$.
| {
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How do I simplify $\frac{\sqrt{21}-5}{2} + \frac{2}{\sqrt{21} - 5}$?
How do I simplify the following equation?
$$\frac{\sqrt{21}-5}{2} + \frac{2}{\sqrt{21} - 5}$$
I have no idea where to start. If I multiply either fraction by its denominator I will still end up with a square root. I know the end result should be $-5$.
| $$\frac{\sqrt{21}-5}{2}+\frac{2}{\sqrt{21}-5}=\frac{\sqrt{21}-5}{2}+\frac{2(\sqrt{21}+5)}{(\sqrt{21}-5)(\sqrt{21}+5)}=\frac{\sqrt{21}-5}{2}+\frac{2\sqrt{21}+10}{-4}=\frac{2\sqrt{21}-10}{4}-\frac{2\sqrt{21}+10}{4}=\frac{2\sqrt{21}-10-2\sqrt{21}-10}{4}=-5$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Dice, balls and boxes probability problem (conditional probability) Problem
Suppose there are two boxes $A$ and $B$ such that $A=\{\text{5 red balls and 3 white balls}\}$, $B=\{\text{1 red ball and 2 white balls}\}.$
A dice is thrown, if the result is $3$ or $6$, a ball from the box $A$ is extracted and then placed it in box $B$ and after that a ball from box $B$ is selected. If the result of the dice is neither $3$ nor $6$, then the inverse procedure takes place (a ball from box $B$ is chosen, then is is thrown in box $A$ and afterwards a ball from box $A$ is selected).
Calculate the probability of both balls being red.
The attempt at a solution
I've denoted $R=\{\text{both balls extracted in each step are red}\}.$, $D_{3,6}=\{\text{the result of the dice is 3 or 6}\}.$ I can calculate $P(R)=P(R|D_{3,6})+P(R|{D_{3,6}}^c).$
Using the formula for conditional probability, I have that $$P(R|D_{3,6})=\frac{P(R \cap D_{3,6})}{P(D_{3,6})},$$ $$P(R|{D_{3,6}}^c)=\frac{P(R \cap {D_{3,6}}^c)}{P({D_{3,6})}^c}.$$
I've calculated $P(R \cap D_{3,6})=\frac{5}{8}\frac{2}{4}$, and $P(D_{3,6})=\frac{2}{6}$.
And $P(R \cap {D_{3,6})}^c)=\frac{1}{3}\frac{6}{9}$, $P({D_{3,6})}^c)=\frac{4}{6}$.
Using these results, we obtain $$P(R)=\dfrac{\frac{5}{8}\frac{2}{4}}{\frac{2}{6}}+\dfrac{\frac{1}{3}\frac{6}{9}}{\frac{4}{6}}.$$
In order to know if my result at least makes sense, I've calculated the actual value of that and it is more than $1$, so there is something wrong with my solution. I am looking for a corrected answer and/or an alternative solution. Thanks in advance.
| You can use
$P(R)=P(R|D_{3,6})P(D_{3,6})+P(R|D_{3,6}^{c})P(D_{3,6}^{c})=\left(\frac{5}{8}\cdot\frac{1}{2}\right)\cdot\frac{1}{3}+\left(\frac{1}{3}\cdot\frac{2}{3}\right)\cdot\frac{2}{3}=\frac{109}{432}$.
| {
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How to solve the quadratic equation $x^2-1=2$?
Solve $x^2-1=2$
I have no idea how to do this can somebody please help me? I have tried working it out and I could never get the answer.
| I feel from the comments that you lack some understanding of equations in general, not just quadratic equation since this problem should be quite simple. Let's try and fix that!
First off, what does it mean to solve the equation $x^2-1=2$? It means to find the value(s) that we can plug into the $x$ such that the equality holds. For instance plugging $x=1$ into the expression on the left hand side yields $x^2-1=1^2-1=0$, so this is not a solution since it does not equal $2$ as we wished. Instead of just guessing for various $x$-values, one often uses operations such as addition, multiplication, etc on both sides of the equality to try and isolate $x$.
Let's try to isolate $x$ in the equation using this method.
$$x^2-1=2$$
We want $x$ to alone on one side so first off we want to get rid of the $-1$, we do this by adding $1$ to each side.
\begin{align}x^2-1+1&=2+1 \implies\\
x^2&=3
\end{align}
Next off we want to get rid of the square. We use the opposite operation, square root:
\begin{align}\sqrt{x^2}&=\sqrt{3} \implies\\
x&=\pm\sqrt{3}
\end{align}
Now $x$ is isolated and we are left with the solution. Either $x=\sqrt{3}$ or $x=-\sqrt{3}$.
Let's do a quick calculation of the other example.
\begin{align}8x^2-200&=0 &\text{ Add 200 to each side}\\
8x^2 &=200 &\text{ Divide by 8}\\
x^2 &= \frac{200}{8} &\text{ Simplify}\\
x^2 &= 25 &\text{ Square root}\\
x &= \pm\sqrt{25}&\text{ Simplify}\\
x &= \pm 5.
\end{align}
And that is the solution.
| {
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$(a_n)$ is bounded $\implies \sum a_n \frac{1}{n^2}$ converges Show that if $(a_n)$ is a bounded sequence, then
$$\sum \frac{1}{n^2} a_n $$ converges
I only could prove that this sequence is bounded, but not its convergence.
Thanks in advance.
| $$V \le a_k \le U\tag{k > 0}$$
$$\frac{1}{k^2}V \le \frac{1}{k^2}a_k \le \frac{1}{k^2}U$$
$$\sum_{k=0}^\infty \frac{1}{k^2}V \le \sum_{k=0}^\infty \frac{1}{k^2}a_k \le \sum_{k=0}^\infty \frac{1}{k^2}U$$
$$\frac{\pi^2}{6}V \le \sum_{k=0}^\infty \frac{1}{k^2}a_k \le \frac{\pi^2}{6}U$$
| {
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What is the limit regarding $a$ What is the limit of :
$$ \lim_{x\to 0} \frac{\sin(ax) - \ln(1-2x)}{e^{ax}-1-2x-2x^{2}}$$
I did this with Maclaurin, because my exam is about solving these with MacLaurin.
Gave $$\lim_{x\to 0} \frac{ax-\frac {(ax)^{2}}{3!} +2x +4x^{2}}{1+ax+(ax)^{2} -1-2x-2x^2} = \frac {a+2}{a-2}$$
The answer should be: $a = 2$, limit is $1/2$,
$ a \neq 2$, limit is $1$
| $\displaystyle\lim_{x\to 0}\frac{\sin(ax)-\ln(1-2x)}{e^{ax}-1-2x-2x^2}=\lim_{x\to 0}\frac{ax-\frac{(ax)^3}{3!}+\cdots-[-2x-\frac{(2x)^2}{2}-\frac{(2x)^3}{3}+\cdots]}{1+ax+\frac{(ax)^2}{2!}+\frac{(ax)^3}{3!}+\cdots-1-2x-2x^2}$
$=\displaystyle\lim_{x\to 0}\frac{(a+2)x+2x^2+(\frac{8}{3}-\frac{a^3}{6})x^3+\cdots}{(a-2)x+(\frac{a^2}{2}-2)x^2+\frac{a^3}{6}x^3+\cdots}=\lim_{x\to 0}\frac{(a+2)+2x+(\frac{8}{3}-\frac{a^3}{6})x^2+\cdots}{(a-2)+(\frac{a^2}{2}-2)x+\frac{a^3}{6}x^2+\cdots}$
$\displaystyle=\begin{cases} \frac{a+2}{a-2} &\mbox {, if } a\ne2 \\ \text{does not exist} &\mbox{, if } a=2\end{cases}$
| {
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Prove $3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\ge xyz(x+y+z)^3$
if $x,y,z$ are positive real numbers,Prove:$$3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\ge xyz(x+y+z)^3$$
Additional info:$\sum_{cyc}$ denotes sums over cyclic permutations of the symbols $x,y,z$. I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done: Writing Left hand side in expanded form:$$LHS=3(\sum_{cyc}x^3y^3+\sum_{cyc}xy^4z+3x^2y^2z^2)$$
Right now I don't have good idea were to start(Is expanding and looking for AM-GM is good?).So any hint is appreciated.
Source : This inequality can be found as Problem 42, page 13, in the book "Old and New inequalities" by Titu Andreescu, where it is attributed to Manlio Marangelli.
| The inequality is equivalent to
$$3\sum x^3y^3 +2\sum xy^4z+3x^2y^2z^2\ge 3\sum(xy^3z^2+x^2y^3z) =3\sum xy^3z(x+z)$$
AM-GM gives
$$x^3y^3+xy^4z+x^2y^2z^2\ge 3 x^2y^3z$$
$$y^3z^3+xy^4z+x^2y^2z^2\ge 3xy^3z^2$$
Summing up we have
$$2\sum x^3y^3+2\sum xy^4z+6x^2y^2z^2\ge 3\sum (x^2y^3z+xy^3z^2),$$
or
$$\sum x^3y^3+\sum xy^4z+3x^2y^2z^2\ge \frac32\sum(x^2y^3z+xy^3z^2).\tag{1}$$
Now, note that
$$x^3+y^3\ge xy^2+yx^2.\tag{2}$$
Multiplying (2) with $xyz$ gives
$$\sum xy^4z\ge \frac12\sum(x^2y^3z+x^3y^2z)\tag{3}.$$
Multiplying (2) with $z^3$ gives
$$\sum (x^3z^3+y^3z^3)\ge \sum(xy^2z^3+x^2yz^3).\tag{4}$$
The inequality follows from (1), (3), and (4).
Note: It may not be a good book to work with after all. And I'm sure there are other non brute-force proofs.
| {
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For What $a$ The Linear Equations Have Sloutions $2x+ay-z=-2$
$x-3z=-3$
$x+2y+az=-1$
I have thought about reducing a matrix so in the end I will have an equation with $a$ then I can determine for which $a$ the are one solution/infinite solutions/no solution, but it does not seems to work
I have row reduced the matrix and got
$$\left(\begin{array}{ccc|c}
1 & 2 & a & -1 \\
0 & -2 & 3-a & 4 \\
0 & a-4 & -a-1& -1
\end{array}\right)$$
the solutions are for one solution: $a=5$,$a\neq 2$ no solution for: $a=-5,a=2$
| I got the Row Reduced Form as:
$ \left( \begin{array}{ccc|c}
1 & 2 & a & -1 \\
0 & 1 & \frac{a+3}{2} & 1 \\
0 & a-5 & \frac{-5(a+1)}{2} & -1 \end{array} \right)$
So the determinant comes as $(-3a+10-a^2)$
For no solution, determinant is zero, so, $(-3a+10-a^2)=0$
$\Rightarrow a=-5, a=2$
| {
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How do I solve $x^5 +x^3+x = y$ for $x$? I understand how to solve quadratics, but I do not know how to approach this question. Could anyone show me a step by step solution expression $x$ in terms of $y$?
The explicit question out of the book is to find $f^{-1}(3)$ for $f(x) = x^5 +x^3+x$
So far I have reduced $x^5 +x^3+x = y$ to $y/x - 3/4 = (x^2 + 1/2)^2$ or $y = x((x^2+1/2)^2 + 3/4)$ but Im still just as lost.
| Solving it for a general $y$ will be difficult, but solving for $y=0$ is more doable.
If $x^5+x^3+x=0$, then because the left hand side is divisible by $x$, either $x=0$ or we can divide by $x$ to get $(x^2)^2+x^2+1=0$. This is a quadratic in $x^2$, so by the quadratic formula, $x^2=\frac{-1\pm \sqrt{-3}}{2}.$
Thus, the five solutions are $x=0$ and $x=\pm\sqrt{\frac{-1\pm \sqrt{-3}}{2}}$.
If we want to show that the function is invertible, then because polynomial functions are continuous, it is enough to show that the function is either always increasing or always decreasing. Because polynomial functions are continuously differentiable and polynomials cannot be locally constant without being completely constant, this is equivalent to showing that the first derivative is always non-negative or always non-positive.
The derivative is $5(x^2)^2+3(x^2)+1$. Because $x^2\geq 0$ for all $x$, the derivative is always positive (in fact, always at least $1$), and therefore the function must be invertible.
| {
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How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$
let $x,y,z>0$, find the minimum of the value
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
I think we can use AM-GM inequality to find it.
$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4]{xy^3}$$
$$3x+z=x+x+x+z\ge 4\sqrt[4]{x^3z}$$
but this is not true,because not all four equalities can hold at once.
This problem is from china middle school test,so I think have without Lagrange methods,so I think this inequality have AM-GM inequality
| Here is another way, using Holder's inequality:
$$\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz} \geqslant \frac{(\sqrt[4]{5y \cdot 5\cdot x\cdot z}+\sqrt[4]{2\cdot2z\cdot3y\cdot3x})^4}{xyz} \\ = (\sqrt5+\sqrt6)^4=241+44\sqrt{30}$$
with equality iff $\dfrac{5y}2=\dfrac5{2z}=\dfrac{x}{3y}=\dfrac{z}{3x} \iff (x, y, z) = \left(1, \sqrt\frac2{15}, \sqrt\frac{15}2 \right)$.
| {
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Evaluating $\lim\limits_{(x,y)\rightarrow(1,1)} \frac {\sin(x) - \sin (y)} {x-y}$ I am taking a calculus exam in less than one week, and I've stumbled upon this expression.
$$\lim\limits_{(x,y)\rightarrow(1,1)} \frac {\sin(x) - \sin (y)} {x-y}$$
Which I know to be cos(1), but I cannot seem to find the inequalities to make an $\epsilon-\delta$ proof of said limit.
I've tried coordinate change and Taylor, to no avail. Whenever I do Taylor(1) or equivalent infinitesimals, any variable I have manages to cancel out. If I don't, the thing just grows and grows...
Is there something I am entirely missing from the start?
| You can use this identity:
$$\sin x - \sin y = 2\sin \frac{x - y}{2}\cos \frac{x + y}{2}.$$
Then your limit becomes:
$$\lim_{(x, y) \to (1,1)} \frac{2\sin \frac{x - y}{2}\cos \frac{x + y}{2}}{x - y} =$$
$$2 \lim_{(x, y) \to (1,1)} \frac{\sin \frac{x - y}{2}\cos \frac{x + y}{2}}{x - y} =$$
$$\frac{1}{2} \cdot2 \lim_{(x, y) \to (1,1)} \frac{\sin \frac{x - y}{2}\cos \frac{x + y}{2}}{\frac{x - y}{2}} =$$
$$\lim_{(x, y) \to (1,1)} \frac{\sin \frac{x - y}{2}}{\frac{x - y}{2}} \cdot \cos \frac{x + y}{2}=$$
$$\lim_{(x, y) \to (1,1)} 1 \cdot \cos \frac{x + y}{2}=$$
$$\lim_{(x, y) \to (1,1)} \cos \frac{2}{2},$$
which exactly yields $\cos 1$.
| {
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The Shortest Distance Between 2 Points On The Earth
Assuming that the earth is a perfect sphere with radius 6378 kilometers, what is the expected straight line distance through the earth (in km) between 2 points that are chosen uniformly on the surface of the earth?
| View the sphere as the surface of revolution of the circle $$x^2+y^2=r^2 \tag 1$$ about the $x$-axis. Differentiating both sides of $(1)$ we get
$$
2x\,dx+2y\,dy=0\quad\text{or}\quad x\,dx+y\,dy=0\quad\text{or}\quad\frac{dy}{dx}=\frac{-x}{y}.
$$
The distance from $(r,0)$ to $(x,y)$ is $\sqrt{(x-r)^2+y^2}$, by the Pythagorean theorem.
The element of arc length is
\begin{align}
& \sqrt{(dx)^2+(dy)^2} \quad\text{(by the Pythagorean theorem)} \\[8pt]
= {} & \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ \ dx \\[8pt]
= {} & \sqrt{1+\frac{x^2}{y^2}}\ \ dx = \frac{\sqrt{y^2+x^2}}{y} \, dx = \frac r y \, dx.
\end{align}
When the graph is revolved about the $x$-axis, the point $(x,y)$ traverses a circle of circumference $2\pi y$. So we get an infinitely narrow strip of length $2\pi y$ and infinitely small width $\dfrac r y\, dx$, hence with area $2\pi y\dfrac r y\,dx=2\pi r\,dx$, and every point on that strip is at the same distance from $(r,0)$, namely
$$
\sqrt{(x-r)^2+y^2} = \sqrt{(x-r)^2+(r^2-x^2)}= \sqrt{2r^2-2rx\ {}}.
$$
So summing all these infinitely small quantities we get
\begin{align}
& \int_{-r}^r\ \overbrace{{}\ \sqrt{2r^2-2rx\ {}}\ {}}^{\text{distance}} \ \overbrace{{}\ 2\pi r\,dx\ {}}^{\text{element of area}} \\[8pt]
= {} & (2r)^{3/2}\pi \int_{-r}^r \sqrt{r-x\,{}}\ dx \\[8pt]
= {} & (2r)^{3/2}\pi \int_{2r}^0 \sqrt{u}\,(-du) \\[8pt]
= {} & (2r)^{3/2}\pi \frac 2 3 (2r)^{3/2} = \frac {16} 3 \pi r^3.
\end{align}
Dividing this by the whole surface area $4\pi r^2$ to get the average, we have
$$
\text{average} = \frac{(16/3)\pi r^3}{4\pi r^2} = \frac{4r}3.
$$
Where we saw $2\pi y\dfrac r y\, dx$, the cancelation of the $y$ seems a bit startling. It was in effect discovered by Archimedes in around 250 BC.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/914792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:
Here's how this question works. To motivate what I'll be doing,
consider \begin{equation*} \dfrac{5}{3} = 1 + \dfrac{2}{3}\text{.}
\end{equation*} This is because when 5 is divided by 3, 3 goes into 5
once (hence the $1$ term) and there is a remainder of $2$ (hence the
$\dfrac{2}{3}$ term). Note the following: every division problem can
be decomposed into an integer (the $1$ in this case) plus a fraction,
with the denominator being what you divide by (the $3$ in this
case).
So, when $n$ is divided by 14, the remainder is 10. This can be
written as \begin{equation*} \dfrac{n}{14} = a + \dfrac{10}{14}
\end{equation*} where $a$ is an integer.
We want to find the remainder when $n$ is divided by 7, which I'll
call $r$. So \begin{equation*} \dfrac{n}{7} = b + \dfrac{r}{7}\text{,}
\end{equation*} where $b$ is an integer.
Here's the key point to notice: notice that \begin{equation*}
\dfrac{n}{7} = \dfrac{2n}{14} = 2\left(\dfrac{n}{14}\right)\text{.}
\end{equation*} This is because $\dfrac{1}{7} = \dfrac{2}{14}$.
Thus, \begin{equation*} \dfrac{n}{7} = 2\left(\dfrac{n}{14}\right) =
2\left(a + \dfrac{10}{14}\right) = 2a + 2\left(\dfrac{10}{14}\right) =
2a + \dfrac{10}{7} = 2a + \dfrac{7}{7} + \dfrac{3}{7} = (2a+1) +
\dfrac{3}{7}\text{.} \end{equation*} So, since $a$ is an integer, $2a
+ 1$ is an integer, which is our $b$ from the original equation. Thus, $r = 3$.
To her, this method was not very intuitive. She did understand the explanation. Are there any suggestions for how I can explain this in another way?
| Colloquially (as I find sometimes works with my daughters at high school)
When we knock out $14$s the remainder is $10$. We knock out $7$s - well the $14$s all go because $14 = 2\times 7$ and we're left with $10=7+3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
How can I accurately compute $\sqrt{x + 2} −\sqrt{x}$ when $x$ is large? How can the values of the function $f(x) = \sqrt{x + 2} −\sqrt{x}$ be computed accurately when $x$ is large?
I have tried using Matllab. I am not able to understand when $x$ will be large.
| If directly computing $\sqrt{x+2}-\sqrt{x}$ is giving you problems, you can instead try using
$\sqrt{x+2}-\sqrt{x} = \dfrac{(\sqrt{x+2}-\sqrt{x})(\sqrt{x+2}+\sqrt{x})}{\sqrt{x+2}+\sqrt{x}} = \dfrac{(x+2)-x}{\sqrt{x+2}+\sqrt{x}} = \dfrac{2}{\sqrt{x+2}+\sqrt{x}}$.
For large $x$, this is approximately $\dfrac{2}{\sqrt{x}+\sqrt{x}} = \dfrac{1}{\sqrt{x}}$.
EDIT: I just tried this for $x = 10^{18}$ using MATLAB R2013b.
For $\sqrt{x+2}-\sqrt{x}$ it gives $0$, but for $\dfrac{2}{\sqrt{x+2}+\sqrt{x}}$, it gives $1.0000 \times 10^{-9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/916402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding $\large\zeta_7\left(\zeta_3\right)^5$ where $\large\zeta_n=\cos{\frac{2\pi}{n}}+i\sin{\frac{2\pi}{n}}$ $\large\zeta_7\left(\zeta_3\right)^5$ where $\large\zeta_n=\cos{\frac{2\pi}{n}}+i\sin{\frac{2\pi}{n}}$
I am having trouble getting a final answer that makes sense to me. Here is what I tried:
$\large\zeta_7\left(\zeta_3\right)^5=\left(\large\zeta_{21}\right)^{3}\left(\zeta_{21}\right)^{35}=\left(\zeta_{21}\right)^{38}=\cos{\frac{76\pi}{21}}+i\sin{\frac{76\pi}{21}}$
Is this an acceptable method/answer? Intuitively, it doesn't feel right.
A second attempt yielded:
$\large\zeta_7=\cos{\frac{2\pi}{7}}+i\sin{\frac{2\pi}{7}}$
$\left(\large\zeta_3\right)^5=\cos{\frac{10\pi}{3}}+i\sin{\frac{10\pi}{3}}$
From here, converting them to the form: $\large e^{\left(\frac{2\pi i}{n}\right)}$
$\large\zeta_7=\large e^{\left(\frac{2\pi i}{7}\right)}$
$\left(\large\zeta_3\right)^5=\large e^{5{\left(\frac{2\pi i}{3}\right)}}$
Multiplying these together: $\large e^{\left(\frac{2\pi i}{7}\right)}\large e^{5{\left(\frac{2\pi i}{3}\right)}}=\large e^{\left(\frac{76\pi i}{21}\right)}$.
This gives an equivalent answer to above which means I am either right, or being tricked into thinking I'm right by my poor math. Any help would be greatly appreciated.
| For this you need to know that $$(A \text{ cis } B) \cdot (C \text{ cis } D) = (AC) \text{ cis } (B + D)$$
And that
$$(A \text{ cis } B)^n = \left(A^n \text{ cis } Bn\right)$$
Your problem is
$$\large\zeta_7\left(\zeta_3\right)^5$$
$$\left(1 \text{ cis } \frac{2\pi}{7}\right)\left(1 \text{ cis } \frac{2\pi}{3}\right)^5$$
$$\left(1 \text{ cis } \frac{2\pi}{7}\right)\left(1^5 \text{ cis } \frac{5\cdot 2\pi}{3}\right)$$
$$\left(1 \cdot 1^5 \right)\text{ cis } \left(\frac{2\pi}{7}+ \frac{5\cdot 2\pi}{3}\right)$$
$$1\text{ cis } \left(2\pi\frac{38}{21}\right)$$
And give that radians repeat every $2\pi$:
$$1\text{ cis } \left(2\pi\frac{17}{21}\right)$$
$$\left(1\text{ cis } \frac{2\pi}{21}\right)^{17}$$
$$\left(\zeta_{21}\right)^{17}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/918859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+C$ Wolfram gives this nice result:
$$\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+\text{constant}$$
I have tried writing $\cos 2x = \cos^2x - \sin^2x $ and doing Weierstrass substitution $\tan (x/2) = t$ but its getting very complicated. Any help/hints ?
| Alternative method (akin to finding an integrating factor by inspection):
$\cos x = \cos (2x - x) = \cos 2x \cos x + \sin 2x \sin x$.
Then the integral becomes:
$$\begin{align}
\int \dfrac{\cos 2x \cos x + \sin 2x \sin x}{\cos^{\frac 3 2} 2x} dx
& = \int \dfrac{\cos x}{\sqrt {\cos 2x}} + \dfrac{\sin x \sin 2x}{\cos^\frac 3 2 2x}dx\\
& = \int \dfrac{d (\sin x)}{\sqrt {\cos 2x}} - \dfrac 1 2 \int \dfrac{\sin x \, d(\cos 2x)}{\cos^\frac 3 2 2x}\\
& = \int \dfrac{du}{\sqrt v} - \dfrac 1 2 \int\dfrac{u\, dv}{v^{3/2}} \qquad [u = \sin x,\ v = \cos x]\\
& = \int d\left( \dfrac{u}{\sqrt v} \right)\\
& = \dfrac{u}{\sqrt v}\\
& = \dfrac{\sin x}{\sqrt {\cos 2x}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show that two expressions are equivalent I am trying to prove a hyperbolic trigonometric identity and I ran into the following expression:
$$\frac{\left (\sqrt{x^2+1}+x \right )^2+1}{2\left ( \sqrt{x^2+1} + x \right )} \quad.$$
This expression is supposed to be equivalent to
$$\sqrt{x^2+1} \quad.$$
I tried to algebraically manipulate the original expression to get the required expression and got to:
$$\frac{\left (\sqrt{x^2+1}+x \right )^2+1}{2\left ( \sqrt{x^2+1} + x \right )} = x+\frac{1}{\sqrt{x^2+1} + x} \quad.$$
but I don't know how I can show that
$$x+\frac{1}{\sqrt{x^2+1} + x} = \sqrt{x^2+1}\quad .$$
| What you got is correct . Then, multiply $$\frac{1}{\sqrt{x^2+1}+x}$$ by $$\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x}(=1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
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