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Why can't I prove this statement by simple induction? Sum of $1/2^1 + \cdots+ n/2^n = x$ I have to prove the following:
$$
\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{2 + n}{2^n}.
$$
I am trying to prove this by simple induction. First, I proved that $P(1)$ holds. It clearly does.
I then assume that $n$ is a positive number $> 1$ and that $P(n)$ holds. Hence,
$$
\frac{1}{2}+\ldots+\frac{n}{2^n}=2-\frac{2 + n}{2^n}.
$$
I now add ${\displaystyle \frac{n + 1}{2\cdot 2^n}}$ to both sides to get the following:
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2\cdot 2^n}=2-\frac{2+n}{2^n}+\frac{n+1}{2\cdot 2^n}.
$$
But after manipulating the right side I get the following:
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{3n + 5}{2^{n+1}}.
$$
This is definitely not right since I should be getting
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{n + 3}{2^{n+1}}=2-\frac{2+(n+1)}{2^{n+1}}
$$
to prove that $P(n)$ implies $P(n+1)$.
What am I doing wrong?
| In RHS, $-\dfrac{2+n}{2^n} + \dfrac{n+1}{2^{n+1}} = \dfrac{-4-2n+n+1}{2^{n+1}} = \dfrac{-(n+3)}{2^{n+1}} = -\dfrac{2+(n+1)}{2^{n+1}}$ which is what you want.
| {
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"timestamp": "2023-03-29T00:00:00",
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What's your favorite proof accessible to a general audience? What math statement with proof do you find most beautiful and elegant, where such is accessible to a general audience, meaning you could state, prove, and explain it to a general audience in roughly $5 \pm\epsilon$ minutes.
Let's define 'general audience' as approximately an average adult with education and experience comparable to someone holding a bachelor's degree in any non science major (e.g. history) from an average North American university.
| Here's a variation on $$\sum_{n=1}^{R} n =\frac{R(R+1)}{2}$$ I stumbled onto this a while back.
shows
$$1+2+3+4+5+6+7 = \frac{1}{2}\cdot b\cdot h + \frac{1}{2}\cdot b$$
$$1+2+3+4+5+6+7 = \frac{1}{2}\cdot 7\cdot 7 + \frac{1}{2}\cdot 7$$
then $$1+2+3+4+5+6+7 = \frac{1}{2}\left( 7\cdot 7 + 7\right)$$
thus $$1+2+3+4+5+6+7 = \frac{7(7+1)}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "121",
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Solve $x^4-3x^2+1=0$ in terms of cosine. I put the equation in the form of a quadratic:
$(x^2)^2-3x^2+1=0$
Then using the quadratic formula,
$x^2=\frac{3\pm\sqrt{9-4}}{2}$
$x^2=\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$
$x=\pm\frac{1+\sqrt{5}}{2}$ and $\pm\frac{1-\sqrt{5}}{2}$
So there are four roots as expected given the equation is a quartic. But I really don't know how to put the answers in terms of cosine. Any hints?
| $\cos(4 t) = 8 \cos(t)^4 - 8 \cos(t)^2 + 1$, so if $x = \sqrt{3} \cos(t)$,
$$x^4 - 3 x^2 = 9 (\cos(t)^4 - \cos(t)^2) = \dfrac{9}{8} (\cos(4t) - 1)$$
Thus to solve $x^4 - 3 x^2 + c = 0$, take $t = \arccos(1 - 8c/9)/4$ and
$$x = \sqrt{3} \cos(t) = \sqrt{3} \cos \left( \dfrac{1}{4} \arccos\left(1 - \dfrac{8c}{9}\right)\right)$$
If you want this solution to be real, you need $1 \ge 1 - 8c/9 \ge -1$, i.e.
$0 \le c \le 9/4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do you solve the equation $ (z^2-1)^2 = 4 ? $ $ (z^2-1)^2 = 4 \iff $$z_1 = 3 $ and $ z_2=-1$
$arg(z_1)= 0 , arg(z_2) = \pi$
$$ z_1 = \sqrt{3} \left(\cos\left(\frac{2\pi k}{2}\right) + i\sin\left(\frac{2\pi k}{2}\right)\right)$$
$$z_2 = i \left(\cos\left(\frac{2\pi k}{2} +\pi\right) + i\sin\left(\frac{2\pi k}{2} +\pi\right)\right) \mid k= \{ 0,1 \}$$
$$ z = \{ \sqrt{3}, -\sqrt{3}, i, -i \}$$
Is this correct?
| $(z^2-1)^2=4\Leftrightarrow (z^2-1)^2-2^2=0\Leftrightarrow (z^2-1-2)(z^2-1+2)=0\Leftrightarrow(z^2-3)(z^2+1)=0\Leftrightarrow (z-\sqrt 3)(z+\sqrt 3)(z-i)(z+i)=0$
It follows that $z\in\{\sqrt 3, -\sqrt 3, i,-i\}$
.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $F^2_{n+1} - F_nF_{n+2} = (-1)^n$ This is a question about Fibonacci sequences, a sequence in which the previous terms build up upon the current term (e.g. $F_1 + F_2 = F_3$ where $F_1 = F_2 = 1$). How would I go about proving $F^2_{n+1} - F_nF_{n+2} = (-1)^n$?
| you can see $$\begin{bmatrix}
f_{n+2} & f_{n+1}\\
f_{n+1}& f_{n}
\end{bmatrix}=(\begin{bmatrix}
1 &1 \\
1 & 0
\end{bmatrix})^{n+1}$$for example put n=1 $$\begin{bmatrix}
f_{3} & f_{2}\\
f_{2}& f_{1}
\end{bmatrix}=(\begin{bmatrix}
1 &1 \\
1 & 0
\end{bmatrix})^{2}=\begin{bmatrix}
1 &1 \\
1 & 0
\end{bmatrix}\begin{bmatrix}
1 &1 \\
1 & 0
\end{bmatrix}=\begin{bmatrix}
2 &1 \\
1 & 1
\end{bmatrix} $$if you put n=2 $$\begin{bmatrix}
f_{4} & f_{3}\\
f_{3}& f_{2}
\end{bmatrix}=(\begin{bmatrix}
1 &1 \\
1 & 0
\end{bmatrix})^{3}=\begin{bmatrix}
2 &1 \\
1 & 0
\end{bmatrix}\begin{bmatrix}
1 &1 \\
1 & 0
\end{bmatrix}=\begin{bmatrix}
3 &2 \\
2 & 1
\end{bmatrix} $$f(1)=1 ,f(2)=1 ,f(3)=2, f(4)=3 and so on ...
so now det(l.h.s)=det(R.h.s) $$det(\begin{bmatrix}
f_{n+2} & f_{n+1}\\
f_{n+1}& f_{n}
\end{bmatrix})=det(\begin{bmatrix}
1 &1 \\
1 & 0
\end{bmatrix})^{n+1}) \\f_{n+2}f_{n}-f_{n+1}f_{n+1}=0-(-1)^{n+1}\\f_{n+2}f_{n}-f_{n+1}f_{n+1}=-(-1)(-1)^{n}=(-1)^{n}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How is $x^2+1=(1/{x^2})[1-{1}/{x^2}+{1}/{x^4}-{1}/{x^6}+\cdots]$? The author of my book writes:
$$x^2+1=x^2\left(1+\frac{1}{x^2}\right)$$
$$=\frac{1}{x^2}\left[1-\frac{1}{x^2}+\frac{1}{x^4}-\frac{1}{x^6}+\cdots\right]$$
I do not understand the last step. How did the author write the last step. Please help.
| I believe your author intended to write
$$
\frac{1}{x^2+1}=\frac{1}{x^2\left(1+\frac{1}{x^2}\right)}
\\
=\frac{1}{x^2}\left[1-\frac{1}{x^2}+\frac{1}{x^4}-\frac{1}{x^6}+\cdots\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Evaluate $\sum_{n=1}^\infty nx^{n-1}$ How can you evaluate $\sum_{n=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}$ without relying on the fact that it's the derivative of $\sum_{n=1}^\infty x^n = \frac{1}{1-x} $?
| another way to prove is :
$$ a=1+2x+3x^2+4x^3+5x^4+6x^5+...\\$$$$lhs=1+x+x^2+x^3+x^4+x^5+x^6+...\\+x+x^2+x^3+x^4+x^5+x^6+...\\+x^2+x^3+x^4+x^5+x^6+...\\+x^3+x^4+x^5+x^6+...\\x^4+x^5+x^6+...\\...\\$$so$$lhs=1+x+x^2+x^3+x^4+x^5+x^6+...=\frac{1}{1-x}\\+x+x^2+x^3+x^4+x^5+x^6+...=\frac{x}{1-x}\\+x^2+x^3+x^4+x^5+x^6+...=\frac{x^2}{1-x}\\+x^3+x^4+x^5+x^6+...=\frac{x^3}{1-x}\\x^4+x^5+x^6+... =+\frac{x^4}{1-x}\\...\\$$factor denominator $$\frac{1}{1-x}+ \frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\frac{x^4}{1-x}+...\\\frac{1}{1-x}(1+x+x^2+x^3+x^4+x^5+x^6+...)=\frac{1}{1-x}*\frac{1}{1-x}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the inverse function $x + \sqrt {x}$ $ Y = x + \sqrt {x} $
Hello , I want to find the inverse function of this function , I know that it's injective How to prove the $f(x) = \sqrt{x + \sqrt{x}}$ is injective. but do not know how to make the inverse function step by step.
| Given $y$, you need to find $x\ge0$ such that $y=x+\sqrt{x}$ or
$$
y-x=\sqrt{x}
$$
A necessary condition is $(y-x)^2=x$, which becomes
$$
y^2-2xy+x^2=x
$$
or
$$
x^2-(2y+1)x+y^2=0
$$
Can you find $x$ from this? Which one of the two solutions have you to consider?
Note that the condition $y-x=\sqrt{x}$ implies $y\ge x$, so the solution you must choose is the only one that satisfies $y\ge x$.
Since you know that the function is injective, only one solution is acceptable; the two roots are $x_1=\dfrac{2y+1+\sqrt{4y+1}}{2}$ and $x_2=\dfrac{2y+1-\sqrt{4y+1}}{2}$. The first one is clearly greater than $y$, so we have $x=\dfrac{2y+1-\sqrt{4y+1}}{2}$, which is acceptable only for $y\ge0$. Indeed from $\dfrac{2y+1-\sqrt{4y+1}}{2}\le y$ we get $2y+1-\sqrt{4y+1}\le 2y$ or $1\le\sqrt{4y+1}$. So the inverse function is $y\mapsto \dfrac{2y+1-\sqrt{4y+1}}{2}$, defined for $y\ge0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to draw a contour map of $ f(x,y)=x^2+y^2+xy$ I have used a program to see that it is an ellipse but I want to know the process of thinking to actually draw the contour map myself.
$x^2+y^2+xy=C$ for $C=0,1,2,3,...$
I can't seem to get it into an ellipse form. What should I do?
Thanks!
| I assume that you have some knowledge of linear algebra.
The quadratic form $x^2 + 2 \frac{1}{2} xy + y^2$ can also be written as a dot product between two vectors
$$x^2 + 2 \frac{1}{2} xy + y^2 = \langle \left (\begin{array}{cc} 1 & 1/2 \\ 1/2 & 1 \end{array} \right ) \cdot \left (\begin{array}{c} x \\ y \end{array} \right ) , \left (\begin{array}{c} x \\ y \end{array} \right ) \rangle= \langle M v, v \rangle$$
Let's look a bit at the matrix of this quadratic form
$$M = \left(\begin{array}{cc} 1 & 1/2 \\ 1/2 & 1 \end{array} \right )$$
It is a real symmetric matrix so it has real eigenvalues and it has a orthonormal basis of eigenvectors. The eigenvalues are $3/2$ and $1/2$ and we also obtain unit eigenvectors:
$$\left(\begin{array}{cc} 1 & 1/2\\ 1/2 & 1 \end{array} \right ) \cdot \left (\begin{array}{c} 1/\sqrt{2}\\ 1/\sqrt{2}\end{array} \right ) = 3/2 \cdot \left(\begin{array}{c} 1/\sqrt{2}\\ 1/\sqrt{2}\end{array} \right )\\
\left(\begin{array}{cc} 1 &1/2\\1/2 & 1 \end{array} \right ) \cdot \left (\begin{array}{c} -1/\sqrt{2}\\ 1/\sqrt{2}\end{array} \right ) = 1/2 \cdot \left(\begin{array}{c} -\frac{1}{\sqrt{2}}\\ 1/\sqrt{2}\end{array} \right )
$$
Therefore we have the equality:
$$\left(\begin{array}{cc} 1 & 1/2\\ 1/2 & 1 \end{array} \right ) \cdot \left(\begin{array}{cc} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{array} \right) =\left(\begin{array}{cc} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{array} \right) \cdot \left(\begin{array}{cc} 3/2 & 0\\ 0 & 1/2 \end{array} \right ) $$
or
$$\left(\begin{array}{cc} 1 & 1/2\\ 1/2 & 1 \end{array} \right ) = \left(\begin{array}{cc} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{array} \right) \cdot \left(\begin{array}{cc} 3/2 & 0\\ 0 & 1/2 \end{array} \right ) \cdot \left(\begin{array}{cc} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{array} \right)^{-1} $$
we wrote the matrix of the quadratic form $M$ as
$$M = R\cdot D \cdot R^{-1}$$
where $D$ is the diagonal matrix $\left(\begin{array}{cc} 3/2 & 0\\ 0 & 1/2 \end{array} \right )$ formed with the eigenvalues and $R$ is the rotation matrix formed with the corresponding unit eigenvectors $R = \left(\begin{array}{cc} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{array} \right) = \left(\begin{array}{cc} \cos\frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos\frac{\pi}{4} \end{array} \right) $
The equation in the vector $v$ giving the curve is
$$\langle Mv, v \rangle = C$$
or
$$\langle R D R^{-1} v, v \rangle = C$$
The rotation matrix $R$ preserves the dot product $\langle \cdot, \cdot \rangle$ so we have $ \langle R D R^{-1} v, v \rangle = \langle D R^{-1} v, R^{-1} v \rangle $
Therefore the equation of the curve is
$$\langle D R^{-1} v, R^{-1} v \rangle = C$$
Denote by $w = R^{-1} v$. We have $\langle D w, w \rangle = C$ and this equation in $w$ give an ellipse in standard form $E_0$. Our curve consists of all $v$ so that $w = R^{-1} v \in E_0$, that is $v \in R E_0$. Therefore, our curve is the rotation of the ellipse $E_0$ by the matrix $R$. Note that $R$ is a rotation of angle $\frac{\pi}{4}$ counterclockwise.
$E_0$ is the ellipse $3/2\cdot x^2 + 1/2\cdot y^2 = C$. Our curve $E$ is the rotation counterclockwise of $E_0$ by angle $\frac{\pi}{4}$
In general, if the matrix of the quadratic form is $M = R_{\alpha} D R_{\alpha}^{-1}$, then the curve
$$\langle Mv, v\rangle = C$$
is the rotation by angle $\alpha$ of the conic
$$\langle D w, w \rangle = C$$
Obs: It's customary that for ellipses in standard form the longer axis is along the $x$-axis. Therefore, it seems preferable to use the decomposition $M = R D R^{-1}$ where the diagonal of $D$ has elements in increasing order. We notice that our curves are also rotations with angle $\frac{\pi}{4}$ clockwise of ellipses $1/2\cdot x^2 + 3/2\cdot y^2 =C$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration by parts and $dx$ notation Please overview this integral evaluation:
$$ \int x^3 \arctan(x^2)dx = \frac{x^4}{4}\arctan(x^2) - \int \frac{1}{1+x^4}2x dx $$
Let's evaluate the right term:
$$\int \frac{1}{1+x^4}\color{Blue}{2x dx} = \int \frac{1}{1+x^4}\color{Blue}{dx^2} = \int y^2 \frac{1}{1+y^2} \color{Red}{dy} = ...$$
Can you explain the legitimacy of those equalities?
Yeah, I know that $(x^2)' = 2x$, but at the middle, we have the term $\frac{1}{1+x^4}$ stands by itself attached to $dx^2$. What's the meaning of that? And at the right we have $dy$ popped out of the blue (Funny, because I colored it blue).
| Hint:
$$\int\frac{2x}{1+x^4}\,dx$$
can be evaluated through partial fraction decomposition, i.e. by noticing that:
$$\frac{1}{1-\sqrt{2}\,x+x^2}-\frac{1}{1+\sqrt{2}\,x+x^2} = \frac{2\sqrt{2}\,x}{1+x^4}.$$
The antiderivatives of both terms in the LHS is an arctangent.
However, by noticing that $x\to\sqrt{x}$ is a smooth bijective map on $\mathbb{R}^+$, a change of variable ($x=\sqrt{u}$, so that $dx=\frac{1}{2\sqrt{u}}\,du$) gives:
$$ \int_{0}^{t}\frac{2x}{1+x^4}\,dx = \int_{0}^{t^2}\frac{du}{1+u^2} = \arctan(t^2).$$
| {
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Last 2 digits of $2345^{369}$ http://i.stack.imgur.com/hte3J.jpg
This webpage says last 2 digits of $2345^{369}$ is $75$.
But considering only last 2 digits:
$45^1 = 45$
$45^2 = 25$
$45^3 = 25$
The last 2 digits are always 25. Aren't they ?
| If we take $n=4A+1=10B+5$ where $A,B$ are arbitrary integers
$4A+1=10B+5\iff4(A-1)=10B\iff2(A-1)=5B\implies\dfrac{5B}2=A-1$ which is an integer
As $(2,5)=1,2$ must divide $B\implies B=2C$(say) for some integer $C$
$\implies n=20C+5$
observe that $(10B+5)^2=100(B^2+B)=25\equiv25\pmod{100} \ \ \ \ (1)$
As $((10B+5)^{n+2},100)=25$ for $n\ge0$ let us consider $(20C+5)^n\pmod{\dfrac{100}{25}}$
As $20c+5\equiv1\pmod4\implies(20c+5)^n\equiv1^n\equiv1$
As $a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod{m\cdot c}$
$\implies(20C+5)^n(20C+5)^2\equiv1\cdot(20C+5)^2\pmod{4(20C+5)^2}$
Using $(1),25|(20C+5)^2\implies100|4(20C+5)^2,$
$(20C+5)^{n+2}\equiv(20C+5)^2\pmod{100}$ for $n\ge0$
But by $(1),(20C+5)^2\equiv25\pmod{100}$
| {
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Help with fraction inequality Let $a,b,c$ be three numbers such that:
*
*$a,b\in (0.5,1)$
*$c \in (0.25,0.5)$
*$c < 0.5a$
*$c > 0.5b$
*$a + b < 1 + c$
Let
$$f(a,b,c) = \frac{1+c}{a+\frac{bc}{c+0.5b}}$$
*
*What is the best upper bound we can give of $f$?
*Can we show that $f\leq 1.5$? (can you find $a,b,c$ such that $f(a,b,c)>1.5$?)
Currently, my best upper bound is $\frac{15}{8}$, but it seems it's not close to being tight (see my proof below) as on one place I use the upper range (in the numerator) of $c$ and on different stage I use the lower range of $c$ (in the denominator).
My first attempt showed $f<2.5$:
$$f(a,b,c) = \frac{1+c}{a+\frac{bc}{c+0.5b}}<\frac{1+0.5a}{a}=0.5+\frac{1}{a}=2.5$$
Found Improvement to $f<\frac{8}{3}$:
$$f(a,b,c) = \frac{1+c}{a+\frac{bc}{c+0.5b}}<\frac{1+c}{a+\frac{0.5b^2}{0.5b+0.5b}}=\frac{1+c}{a+0.5b}<\frac{1+0.5a}{a+0.5b}<\frac{1+1}{0.5+0.5^2}=\frac{8}{3}$$
Further improvement to $f<\frac{15}{8}$ (using $c<b$ which is obvious by their domains):
$$f(a,b,c) = \frac{1+c}{a+\frac{bc}{c+0.5b}}<\frac{1+c}{2c+\frac{c^2}{c+0.5c}}$$
$$=\frac{1+c}{2c+\frac{2}{3}c}=\frac{1+c}{\frac{8}{3}c}=\frac{3}{8}+\frac{1}{\frac{8}{3}c}$$
$$=\frac{3}{8}+\frac{3}{8c}<\frac{3}{8}+\frac{12}{8}=\frac{15}{8}$$
| We have $\frac12 < b < 2c < a < 1$. Hence:
$$\frac{1+c}{a+\frac{bc}{c+b/2}} < \frac{1+c}{a+\frac{bc}{2c}} < \frac{1+\frac12a}{\frac32a} = \frac2{3a}+\frac13 < \frac43+\frac13 = \frac53$$
Further as $b, 2c, a \to \frac12$, the the function comes arbitrarily close to $\frac53$, so this is indeed the best/lowest upper bound.
You already have cases where $f> 1.5$, e.g. $a=0.52, b = 0.51, c = \frac12(a+b)$...
| {
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Induction proof concerning Pell numbers Problem: The Pell numbers $p_n$ are defined by the recurrence relation \begin{align*} p_{n+1} = 2p_n + p_{n-1} \end{align*} for $n \geq 1$, together with $p_0 = 0$ and $p_1 = 1$.
Prove with mathematical induction that \begin{align*} p_{n+1} p_{n-1} - p_n^2 = (-1)^n \end{align*} for every $n \in \mathbb{N} \setminus \left\{0\right\}$.
Proof: Initial step: for $n = 1$ we have $p_2 p_0 - p_1^2 = (-1)$ which is true given the initial conditions.
Inductive step: Suppose the above expression is true for $n > 1$. Then we have to show that it is also true for $n + 1$. So we have \begin{align*} p_{n+1} p_{n-1} - p_n^2 = (-1)^n. \end{align*} I now plugged in the expression for $p_{n+1}$ given above, and thus got: \begin{align*} (2p_n + p_{n-1}) p_{n-1} - p_n^2 = (-1)^n, \end{align*} or \begin{align*} 2p_n p_{n-1} + p_{n-1}^2 - p_n^2 = (-1)^n. \end{align*}
I'm not sure if I'm doing this right, and I don't know how to proceed. Any help would be appreciated.
| Look at what you want to show: you want to show that $p_{n+2}p_n-p_{n+1}^2=(-1)^{n+1}$. You know that $p_{n+2}=2p_{n+1}+p_n$, so this is equivalent to
$$(-1)^{n+1}=(2p_{n+1}+p_n)p_n-p_{n+1}^2=p_n^2+2p_np_{n+1}-p_{n+1}^2\;.\tag{1}$$
Your induction hypothesis is that $p_{n+1}p_{n-1}-p_n^2=(-1)^n$; that has a $p_{n-1}$ that isn’t present in $(1)$, so perhaps we should manipulate the target $(1)$ a bit more to get a $p_{n-1}$. We might apply $p_{n+1}=2p_n+p_{n-1}$ to one of the factors of $p_{n+1}$ in $p_{n+1}^2$ to get
$$\begin{align*}
p_n^2+2p_{n+1}-p_{n+1}^2&=p_n^2+2p_{n+1}-p_{n+1}(2p_n+p_{n-1})\\
&=p_n^2+2p_np_{n+1}-2p_np_{n+1}-p_{n-1}p_{n+1}\\
&=p_n^2-p_{n-1}p_{n+1}\;.
\end{align*}$$
Can you finish it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:
*
*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
*$f(0)=0$ and $f(1)=0$;
*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ and $x=1$ is equal to $1$.
Compute the arc length, $L$, for the function $f$. The goal is to minimize $L$ given the three conditions above.
$\mathbf{\color{red}{\text{Contest results:}}}$
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline
\text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\
\text{2} & \text{Glen O} & {} & {} & 2.78567 \\
\text{3} & \text{mickep} & {} & {} & 2.81108 \\
\text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\
\text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline
\text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\
\text{-} & \text{Narasimham} & {} & {} & 2.78 \\
\end{array}$$
Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\
\text{6} & -4x\ln x & {} & {} & 3.21360 \\
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\
\end{array}$$
| The most recent overkill contribution - which is at least somewhat interesting, though not the optimal - is the following behemoth: $f(x)=-0.00010156(b^{x-1/2}+b^{-(x-1/2)})+1.120357$ where $b=121716670.4$ You can think of this as a hyperbolic cosine with a different base. This has arc length $2.8788364$. The above answer does beat polynomials of the form $p(x)=a(x-1/2)^n+b$. Such polynomials can do no better than 2.8940115.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 12,
"answer_id": 11
} |
How to evaluate the limit $\lim_\limits{x\to 0+ } \frac{1}{\sqrt{x}}\left ( \frac{1}{\sin x} - \frac{1}{x}\right )$? $$\lim_{x\to 0^+ } \frac{1}{\sqrt{x}}\left ( \frac{1}{\sin x} - \frac{1}{x}\right ) =\ ?$$
I rearranged it as
$$\lim_{x\to 0^+ } \frac{x-\sin x}{x\sqrt{x}\sin x} = \lim_{x\to0^+ } \frac{x-\sin x}{x^{\frac{3}{2}}\sin x}$$
Which gives an indetermination of the form $0/0$. Then, I tried L'Hospital:
$$\lim_{x\to 0^+ } \frac{x-\sin x}{x^{\frac{3}{2}}\sin x} = \lim_{x\to 0^+ } \frac{1-\cos x}{\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}} \cos x}$$
Should I continue to apply L'Hospital or is there a simpler way to solve it?
| Write it as
$$ \frac{x-\sin x}{x^{3/2}\sin x} \sim_{x\sim 0} \frac{x^3/3!}{x^{3/2}\,x} = \frac{1}{3!}x^{1/2}\longrightarrow_{x\to 0}0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Limit problems and quandaries: finding $\lim_\limits{n\to \infty } {({n^2-n\over n^2+1})^{n+10} }$. Find $\lim_\limits{n\to \infty } {({n^2-n\over n^2+1})^{n+10} }$. What I did is:
$\lim_\limits{n\to \infty }{({n^2-n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty } {({n^2+1-1-n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty }{(1-{1+n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty}[{(1-{1+n\over n^2+1})^{ n^2+1\over 1+n}}]^{(n+10)(1+n)\over (n^2+1)}$. Denoting: ${ n^2+1\over 1+n}=t$ ($t\to \infty$ as $n\to \infty$) we get: $\lim_\limits{n\to \infty}[{(1+{(-1)\over t})^{ t}}]^{(n+10)(1+n)\over (n^2+1)}=\lim_\limits{n\to \infty}e^{-{(n+10)(1+n)\over (n^2+1)}}=e^{\lim_\limits{n\to \infty}-{(n+10)(1+n)\over (n^2+1)}}=e^{-1}$.
My question is: how can I know it is defined and lawful? I know I can use the continuity of $e$, but I don't know if it is okay that I separated the limits that way(Because maybe the first limit is not even $e$ considering other things)
.I Would appreciate your help...
| It is much simpler to take logs. We need to use the following theorem for that purpose.
If $a_{n} > 0$ then $\lim_{n \to \infty}a_{n}$ exists if and only if $\lim_{n \to \infty}\log a_{n}$ exists or is $-\infty$. If $\lim_{n \to \infty}\log a_{n} = A$ then $\lim_{n \to \infty}a_{n} = e^{A}$ and if $\log a_{n} \to -\infty$ then $a_{n} \to 0$.
Supposing then that the desired limit is $L$, we have $$\begin{aligned}\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{n^{2} - n}{n^{2} + 1}\right)^{n + 10}\right\}\\
&= \lim_{n \to \infty}\log\left(\frac{n^{2} - n}{n^{2} + 1}\right)^{n + 10}\text{ (by continuity of log)}\\
&= \lim_{n \to \infty}(n + 10)\log\left(\frac{n^{2} - n}{n^{2} + 1}\right)\\
&= \lim_{n \to \infty}n\log\left(\frac{n^{2} - n}{n^{2} + 1}\right) + 10\cdot\log\left(\frac{n^{2} - n}{n^{2} + 1}\right)\\
&= \lim_{n \to \infty}n\log\left(\frac{n^{2} - n}{n^{2} + 1}\right) + 10\cdot\log 1\\
&= \lim_{n \to \infty}n\log\left(1 - \frac{n + 1}{n^{2} + 1}\right)\\
&= \lim_{n \to \infty}n\left(-\dfrac{n + 1}{n^{2} + 1}\right)\dfrac{\log\left(1 - \dfrac{n + 1}{n^{2} + 1}\right)}{\left(-\dfrac{n + 1}{n^{2} + 1}\right)}\\
&= \lim_{n \to \infty}n\left(-\dfrac{n + 1}{n^{2} + 1}\right)\cdot\lim_{x \to 0}\frac{\log(1 + x)}{x}\text{ (by putting }x = -(n + 1)/(n^{2} + 1))\\
&= -\lim_{n \to \infty}\dfrac{n^{2} + n}{n^{2} + 1}\\
&= -\lim_{n \to \infty}\dfrac{1 + \dfrac{1}{n}}{1 + \dfrac{1}{n^{2}}}\\
&= -1\end{aligned}$$ Hence $L = e^{-1} = 1/e$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Find $\lim_{n\to \infty}({1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+\cdots+{1\over \sqrt{n^2+n}})$ strong textFind $\lim_\limits{n\to \infty}\left({1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+\cdots+{1\over \sqrt{n^2+n}}\right)$.
I do know it is bounded by $1$. I tried using the sandwich rule with no success. How can I solve it?
| Note
$$\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+n}}\le\lim_\limits{n\to \infty}\left({1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+\cdots+{1\over \sqrt{n^2+n}}\right)\le\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+1}}$$
Since
$$\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+n}}=\lim_\limits{n\to \infty}\frac{1}{\sqrt{1+\frac{1}{n}}}=1$$
and
$$\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+1}}=\lim_\limits{n\to \infty}\frac{1}{\sqrt{1+\frac{1}{n^2}}}=1$$
we have that the limit of the original is $1$ by the sandwich rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Solve $(\alpha,\beta)$ for $\lim_{n\to\infty} \frac{\sqrt[n^2]{1!2!\cdots n!}}{n^\alpha} = \beta$ Find the ordered pair $(\alpha,\beta)$ with non-infinite $\beta \ne 0$ such that $$\lim_{n\to\infty} \frac{\sqrt[n^2]{1!2!\cdots n!}}{n^\alpha} = \beta$$
My approach:
$$\ln (1!2!\cdots n!) = (n)\ln 1 + (n-1)\ln 2 + \cdots + (2)\ln (n-1) + \ln(n) \\ \begin{align} = n\ln\left(\frac{1}{n}\right) + (n-1)\ln\left(\frac{2}{n}\right) + \cdots + \ln\left(\frac{n}{n}\right) + \frac{(n)(n+1)}{2} \ln (n)\end{align}$$
Then $$\ln(\sqrt[n^2]{1!2!\cdots n!}) = \frac{1}{n^2} \ln(1!2!\cdots n!) = \frac{n+1}{n} \cdot \ln n + \frac{1}{n} \left[\sum_{m=1}^n \left(\frac{n+1-m}{n} \cdot \ln \frac{m}{n}\right)\right]$$
And that's about as far as I got. Any ideas about proceeding with this method or perhaps even with a different method?
Thanks
A
| Another way: by Stolz-Cesaro theorem, we that
$$\begin{align}
L&=\lim_{n\to\infty} \frac{1}{n^{2}} \left(\sum_{k=1}^{n} \log(k!)-\alpha n^2\log(n) \right)\\
&=\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}\\
&=\lim_{n\to\infty} \frac{1}{n^2-(n-1)^2} \left(\log (n!) - \alpha n^2\log(n)+\alpha(n-1)^2\log(n-1)\right)\\
&=\lim_{n\to\infty} \frac{1}{2n-1} \left(\log (n!) -\alpha n^2\log(n)+\alpha(n-1)^2(\log(n)+\log(1-1/n))\right)\\
&=\lim_{n\to\infty} \frac{1}{2n-1} \left(n\log(n) -n+o(n) -2\alpha n\log(n)+\alpha(n-1)^2\Big(-\frac{1}{n}+o(1/n)\Big)\right)\\
&=\lim_{n\to\infty} \frac{1}{2n-1} \left((1-2\alpha)n\log(n) -(1+\alpha)n +o(n)\right)\\
&=\begin{cases}
\text{sgn}(1-2\alpha)\cdot +\infty &\text{if $\alpha\not=1/2$,}\\
-\frac{3}{4}&\text{if $\alpha=1/2$,}
\end{cases}
\end{align}$$
where we applied the Stirling approximation $\log (n !)=n\log n -n+o(n)$.
Therefore we may conclude that
$$\lim_{n\to\infty} \frac{\sqrt[n^2]{1!2!\cdots n!}}{n^\alpha} =e^L.$$
Hence $\alpha=1/2$ and $\beta=e^{-3/4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Is there another way to prove $(x-n)^2 = (n-x)^2$ Let's say $n$ is $4$. So, I came up with the solution below.
*
*$(x-4)^2 = (x-4)(x-4) = x^2 - 8x + 16$
*$(4-x)^2 = (4-x)(4-x) = 16 - 8x + x^2 = x^2 - 8x + 16$
I was wondering if there is another way to proof that $(x-n)^2$ equals to $(n-x)^2$ ?
| You can argue like this:
$$(x-n)^2 = ((-1)(n-x))^2 = (-1)^2 (n-x)^2 = (n-x)^2.$$
| {
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"url": "https://math.stackexchange.com/questions/1125140",
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"source": "stackexchange",
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If $a,b,c\in\mathbb{R^+}$ such that $ abc = 1 $ and $ ab + bc + ca = 5 $. Prove that $ 17/4 \leq (a+b+c)\leq 1+ \sqrt{32}. $ If $a,b,c\in\mathbb{R^+}$ such that $ abc = 1 $ and $ ab + bc + ca = 5 $. Prove that
$$
\frac{17}{4} \leq (a+b+c)\leq 1+ \sqrt{32}.
$$
My attempt
Tried using Vieta but it didn't work. Also I used some standard inequalities but got $ \geq\sqrt{15} $.
| Let $P(x)=(x-a)(x-b)(x-c)$.
$$P(x)=x^3-sx^2+5x-1$$
with $s=a+b+c$
$P$ must have 3 reals roots. So discrimant of $P$ must be positive. From the formula of discriminant, we get
$$\Delta(P)=-4s^3+25s^2+90s-527=R(s)$$
But the polynomial $R$ has an easy root : $\frac{17}{4}$. So we factorize :
$$R(s)=-4\left(s-\frac{17}{4}\right)(s^2-2s-31)=-4\left(s-\frac{17}{4}\right)\left(s-(1+\sqrt{32})\right)\left(s-(1-\sqrt{32})\right)$$
So $R$ is only positive between its two positive roots, so
$$\frac{17}{4}\le s\le 1+\sqrt{32}$$
Note : $R$ is also positive before its negative root, but it would imply $s<0$.
Another solution : The limit cases where $P$ has 3 real roots are when $P$ has two identical roots. So just suppose $b=c$.
Hence $ab^2=1$ and $2ab+b^2=5$ so $Q(b)=b^3-5b+2=0$
But $Q$ has some easy roots : $2$, and $-1\pm\sqrt{2}$.
Use the two positive roots as values for $b$, and deduce $a$ and $a+b+c$, you will find again $\frac{17}{4}$ and $1+\sqrt{32}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Evaluating the Definite Integral $\int_0^{\pi}\cos^{2n} \theta d\theta$ $$\int_0^{\pi}\cos^{2n} \theta d\theta$$
$$u=\cos \theta \implies du= -\sin \theta d\theta \implies d\theta= -\frac{du}{1-u^2} $$
$$\int_{-1}^1 \frac{u^n}{1-u^2} du=\int_{-1}^1 \frac{u^n}{(1-u)(1+u)}du$$
I have no idea what to do next, any guidance is appreciated!
| You have made an error in your substitution. So let us start again from the beginning. I will assume $n$ is a positive integer.
In the integral we let $u = \cos \theta, du = - \sin \theta \, d\theta$ and this leads to
$$d\theta = -\frac{du}{\sqrt{1 - u^2}},$$
while for the limits of integration $(0,\pi) \mapsto (1,-1)$. So
\begin{align*}
\int_0^\pi \cos^{2n} \theta \, d\theta &= \int_{-1}^1 \frac{u^{2n}}{\sqrt{1 - u^2}} \, du\\
&= 2 \int_0^1 \frac{u^{2n}}{\sqrt{1 - u^2}} \, du,
\end{align*}
as the integrand is even between symmetric limits. Now enforcing a substitution of $u \mapsto \sqrt{u}$ yields
\begin{align*}
\int_0^\pi \cos^{2n} \theta \, d\theta &= \int_0^1 \frac{u^{n - \frac{1}{2}}}{\sqrt{1 - u}} \, du\\
&= \int_0^1 u^{(n + \frac{1}{2}) - 1} (1 - u)^{\frac{1}{2} - 1} \, du\\
&= \text{B} \left (n + \frac{1}{2}, \frac{1}{2} \right )\\
&= \frac{\Gamma \left (n + \frac{1}{2} \right ) \Gamma \left (\frac{1}{2} \right )}{\Gamma (n + 1)}.
\end{align*}
As
$$\Gamma \left (\frac{1}{2} \right ) = \sqrt{\pi}, \quad \Gamma (n + 1) = n!, \quad \Gamma \left (n + \frac{1}{2} \right ) = \frac{(2n)! \sqrt{\pi}}{2^{2n} n!}.$$
(a proof for this last result can be found here) we have
$$\int_0^\pi \cos^{2n} \theta \, d\theta = \frac{(2n)! \pi}{2^{2n} (n!)^2} = \frac{\pi}{2^{2n}} \binom{2n}{n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Proving the inequality $\frac12\frac34....\frac{2n-1}{2n}<\frac1{\sqrt{2n+1}}$ How to show this inequality:
$\dfrac{1}{2}\dfrac{3}{4}....\dfrac{2n-1}{2n}<\dfrac{1}{\sqrt{2n+1}}$
Using induction the inequality is verified for $n=1$
now assume that that the inequality holds for $n$,to show it for $n+1$
Then
$\dfrac{1}{2}.\dfrac{3}{4}....\dfrac{2n-1}{2n}\dfrac{2n+1}{2(n+1)}<\dfrac{1}{\sqrt{2n+1}}.\dfrac{2n+1}{2(n+1)}=\dfrac{\sqrt{2n+1}}{2(n+1)}<\dfrac{\sqrt{2(n+1)}}{2(n+1)}=\dfrac{1}{\sqrt{2(n+1)}}$
but I have to make it less than $\dfrac{1}{\sqrt{2n+3}}$ which is not coming.Any help
| Another posibility is to use the fact that $$\frac{k-1}{k}<\frac{k}{k+1} $$ for any k greater or equal to 1. If you write this inequality for k=3,5,...,2n-1 and multiply them, the answer should occur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Evaluate this limit $\lim_{x\to - \infty}\frac{4^{x+3}-3^{x+2}-2^{x+1}}{4^{x+1}+3^{x+2}+2^{x+3}}$ Evaluate the limit
$$\lim_{x\to - \infty}\frac{4^{x+3}-3^{x+2}-2^{x+1}}{4^{x+1}+3^{x+2}+2^{x+3}}.$$
I've tried and it always comes out $\frac{0}{0}$, and l'Hopital doesn't seem to help me much here, what would you do?
| We have
$\begin{align}
\lim_{x\to - \infty}\frac{4^{x+3}-3^{x+2}-2^{x+1}}{4^{x+1}+3^{x+2}+2^{x+3}}&=\lim_{x\to - \infty}\frac{2^{x+1}\Big(4^2(\frac{4}{2})^{x+1}-3(\frac{3}{2})^{x+1}-1\Big)}{2^{x+3}\Big((\frac{1}{2})^2(\frac{4}{2})^{x+1}+\frac{1}{2}(\frac{3}{2})^{x+1}+1\Big)}\\
&=\frac14\lim_{x\to - \infty}\frac{4^2(\frac{4}{2})^{x+1}-3(\frac{3}{2})^{x+1}-1}{(\frac{1}{2})^2(\frac{4}{2})^{x+1}+\frac{1}{2}(\frac{3}{2})^{x+1}+1}\\
&=\frac14\lim_{x\to - \infty}\frac{4^2(\frac{2}{4})^{-x-1}-3(\frac{2}{3})^{-x-1}-1}{(\frac{1}{2})^2(\frac{2}{4})^{-x-1}+\frac{1}{2}(\frac{2}{3})^{-x-1}+1}\\
&=\frac14\lim_{x\to \infty}\frac{4^2(\frac{2}{4})^{x-1}-3(\frac{2}{3})^{x-1}-1}{(\frac{1}{2})^2(\frac{2}{4})^{x-1}+\frac{1}{2}(\frac{2}{3})^{x-1}+1}\\
&=\frac{1}{4} \times (-1)=-\frac{1}{4}.
\end{align}$
where I have change the variable from $-x$ to $x$ at last.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Finding product without working it out What's the easy way to find the solution for the below problem without actually multiplying the numbers:
$$(24 * 24) - (16 * 16)?$$
I tried multiplying the numbers but that is a long way.
The answer is in the below format
A) $(8 \times 8)$
B) $(40 \times 40)$
C) $(8 \times 2)$
D) $(8 \times 40)$
| You can also try and use prime factorization:
\begin{align}(24 \cdot 24) - (16 \cdot 16) &= 24^2 - 16^2 \\ &= (3\cdot 2^3)^2 - (2^4)^2 \\ &= (3^2 \cdot2^6) - (2^8) \\ &= 2^6 \bigg[3^2 - 2^2 \bigg] \\ &= 2^6 \bigg[9 - 4 \bigg] \\ &= 2^6 (5) \\ &= 64(5) \\ &= 320\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Rule for factoring out variable in limit I am a bit confused about the proper way to calculate the limit of a function by factoring out the variable that strives for infinity.
If you have this:
$\lim_{x\rightarrow \infty } \sqrt{x} \left(\sqrt{x+1}-\sqrt{x}\right)$
for some reason you cannot do this:
$ = \lim_{x\rightarrow \infty } \sqrt{x} \left(\sqrt{x(1+ \dfrac{1}{x})}-\sqrt{x}\right) = \lim_{x\rightarrow \infty } \sqrt{x} \left(\sqrt{x}-\sqrt{x}\right) = 0$
but rather should do this:
$ = \lim_{x\rightarrow \infty } \sqrt{x} \frac{\left(\sqrt{x+1}-\sqrt{x}\right)\left(\sqrt{x+1}+\sqrt{x}\right)}{\left(\sqrt{x+1}+\sqrt{x}\right)} = \lim_{x\rightarrow \infty } \sqrt{x} \frac{x+1-x}{\left(\sqrt{x+1}+\sqrt{x}\right)} = \lim_{x\rightarrow \infty } \frac{\sqrt{x}}{\sqrt{x(1+\frac{1}{x})}+\sqrt{x}} = \lim_{x\rightarrow \infty } \frac{\sqrt{x}}{2\sqrt{x}} = \frac{1}{2}$
Why is it legit to factor out $x$ in the second approach but not in the first one? Is there a general rule for when you are allowed to do so and when not?
| Although $\lim_{x \to \infty} \sqrt{x} (3-3) = \lim_{x \to \infty} \sqrt{x} \cdot 0 =0$, in your case you have $\lim_{x \to \infty} = \sqrt{x}(\sqrt{x} - \sqrt{x}) \neq \lim_x \sqrt{x} \cdot (\lim_x \sqrt{x} - \sqrt{x})$ because $\infty \cdot 0 $ is an indeterminate form. Also $\lim_x (\sqrt{x} - \sqrt{x}) \neq \lim_x \sqrt{x} - \lim_x \sqrt{x}$ because \infty - \infty is an indeterminate form too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How To Find The Length Of An Irregular Arc How would you find out the length of an irregular arc. e.g. An arc with a base length of $10$cm and a height of $5$cm - what would be the length of that arc? Is there a specific formula I could use?
| I assume you mean a circular arc, with chord length ("base length") of $w$ and distance along the perpendicular bisector of the base from the base chord to the arc of $h$ (the "height" of the arc).
Although by those definitions, the arc you describe is easy (it is in fact a semi-circle of a circle with radius 5 cm, so has a length of $5\pi$) the general case is not as easy but also tractable using elementary trigonometry:
Call the chord (the base) $AB$, and let $P$ be the midpoint of $AB$, $Q$ the meeting point on the arc, and $O$ be the circle center of the arc from $A$ to $B$. Then $OA = OQ$ and in right triangle $OPA$,
$$
OP = OA - h \\
PA = w/2 \\
(OA)^2 = (OP)^2+(PA)^2 = (OA)^2-2h(OA) +h^2 + \frac{w^2}{4} \\
2h(OA) = h^2 + \frac{w^2}{4}\\
OA = \frac{h}{2} + \frac{w^2}{8h}
\\
\cos \angle POA = \frac{OA-h}{OA} = \frac{ \frac{w^2}{8h} - \frac{h}{2} }
{ \frac{w^2}{8h}+\frac{h}{2} }
$$
and the length of the arc from $A$ to $Q$ and then from $Q$ to $B$ is
$$
2 (OA) \cos^{-1}\frac{w^2-4h^2}{w^2+4h^2} = 2 \frac{4h^2+w^2}{8h}\cos^{-1}\frac{w^2-4h^2}{w^2+4h^2}
$$
In the example given, $w = 10, h=5$ and the arc length is
$$
2 \frac{4\cdot 25 +100}{8\cdot 5}\cos^{-1}\frac{100-4\cdot 25}{100-4\cdot 25}
=2\cdot \frac{200}{40} \cos^{-1}(0) = 2\cdot 5 \cdot \frac{\pi}{2} = 5\pi
$$
If, instead, the height were 3 cm then the circular arc length would be
$$
\frac{34}{3}\cos^{-1}\frac{8}{17} \approx 12.25$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimal polynomial of a matrix satisfying $A^t=A^2$
Let $A\in M_n(\mathbb R)$ be a non-zero non-identity matrix such that $A^t=A^2$. Then is it possible to find the minimal polynomial of $A$?
My try is: the given condition implies that $A^4=A$, hence the minimal polynomial $p(t)$ of $A$ divides $t^4-t=t(t-1)(t^2+t+1)$. So $p(t)$ can be some factor of $t(t-1)(t^2+t+1)$.
Also if $\lambda$ is an eigenvalue of $A$ and $x$ is the corresponding eigenvector (considered as a column vector), then $Ax=\lambda x \implies \lambda x^t=x^t A^t\implies \lambda x^t=x^tA^2\implies \lambda(x^t x)=x^t(A^2x)\implies \lambda(x^t x)=\lambda^2(x^tx)\implies (\lambda^2-\lambda)(x^tx)=0\implies\lambda^2-\lambda=0,\text{ as } x^tx \text{ is non-zero}.$
Hence $0$ and $1$ are the only eigenvalues of $A$. So the minimal polynomial is of the form $p(t)=t(t-1).$ Is this correct or did I misunderstood something?
P.S. $A^t$ denotes the transpose of the matrix $A$.
| Since
$$
A^4=\left(A^2\right)^2=\left(A^T\right)^2=\left(A^2\right)^T=\left(A^T\right)^T=A
$$
the minimal polynomial must divide $x^4-x=x(x-1)(x^2+x+1)$.
For example,
$$
A=\begin{bmatrix}
-\frac12&\frac{\sqrt3}2\\
-\frac{\sqrt3}2&-\frac12
\end{bmatrix}
$$
has minimal polynomial $x^2+x+1$ and $A^2=A^T$.
$$
B=\begin{bmatrix}
-\frac12&\frac{\sqrt3}2&0\\
-\frac{\sqrt3}2&-\frac12&0\\
0&0&1
\end{bmatrix}
$$
has minimal polynomial $x^3-1$ and $B^2=B^T$.
$$
C=\begin{bmatrix}
-\frac12&\frac{\sqrt3}2&0\\
-\frac{\sqrt3}2&-\frac12&0\\
0&0&0
\end{bmatrix}
$$
has minimal polynomial $x^3+x^2+x$ and $C^2=C^T$.
$$
D=\begin{bmatrix}
-\frac12&\frac{\sqrt3}2&0&0\\
-\frac{\sqrt3}2&-\frac12&0&0\\
0&0&1&0\\
0&0&0&0
\end{bmatrix}
$$
has minimal polynomial $x^4-x$ and $D^2=D^T$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1136247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the limit of the sequence $a_n=(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdots(1-\frac{1}{n^2})$? I am stuck on finding the limit of
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdots\left(1-\frac{1}{n^2}\right)$$
Can anybody help?
UPDATE
Is there a solution involving making the fractions like
$$(\frac{n^2-1}{n^2})$$
and using the formula
$$(2^23^24^2...n^2) = (n!)^2$$?
| Hint: The function $sin(x)$ has Zeros at $..., - \pi, 0, \pi, 2 \pi, ...$. Therefore one can write:
$sin(\pi x) = ...(x+2 )(x+1)(x)(x-1)(x-2 )... = ...(x^2-2^2 )x(x^2-1^2)...$.
Rearrange this Expression by dividing through $1^22^2...$ and then Substitute $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
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second derivative with a fraction Say i have $\frac{x^2+a}{x+a}$
the first derivative is rather easy:
$$
\frac{(x+a)(2x)-(x^2+a)(1)}{(x+a)^2}
$$
which is
$$
\frac{2ax - a + x^2}{ (x+a)^2}
$$
But i can't figure out how to do the second derivative correctly. I tried :
$$
\frac{(x+a)(x+a)(2a+2x)-(2ax-a+x^2)(2)(x+a)}{(x+a)^4}
$$
And then divide $(x+a)$ to get
$$
\frac{(x+a)(2a+2x)-(2ax-a+x^2)(2)(x+a)}{(x+a)^3}
$$
But it didn't work out
| it is easier to find f' and f'' by simplifying ,like this $$f(x)=\frac{x^2+a}{x+a}=\\\frac{x^2-a^2+a^2+a}{x+a}=\\\frac{(x-a)(x+a)+(a^2+a)}{x+a}=\\x-a+\frac{a^2+a}{x+a}=\\x-a+(a^2+a)\frac{1}{x+a}\\$$$$f'=1+(a^2+a)\frac{-1}{(x+a)^2}\\f''=0+(a^2+a)\frac{+2(x+a)}{(x+a)^4}=\\f''=(a^2+a)\frac{2}{(x+a)^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1139475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof: $2\sqrt{m}-2 < \sum\limits_{n=1}^m\frac{1}{\sqrt{n}}< 2\sqrt{m}-1$ I know that problems similar to this one, involving either one of the two bounds, have been posted before, but I would like just a hint in the last part of the proof involving the upper bound, with which I have some troubles with being ''convincing'' enough.
Prove that $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$ if $n\geq 1$. Then use this to prove that
$$2\sqrt{m}-2 < \sum_{n=1}^m\frac{1}{\sqrt{n}}< 2\sqrt{m}-1 \qquad \text{if }\ m \geq 2$$
With some help I received here I got the following proof for the first set of inequalities:
Proof (direct):
Since
$$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{1}{2\sqrt{n}}$$
then we have
$$2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}}\qquad (1)$$
Likewise, since
$$\frac{1}{2\sqrt{n}} = \frac{1}{\sqrt{n} + \sqrt{n}} < \frac{1}{\sqrt{n} + \sqrt{n-1}} = \sqrt{n} - \sqrt{n-1}$$
then
$$\frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})\qquad (2)$$
as asserted.
For the second part I tried to prove the lower bound firt:
Proof (direct): From $(1)$ we have
$$\sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}}$$
By summing both sides we get
$$\sum_{n=1}^{m}(\sqrt{n+1} - \sqrt{n}) < \sum_{n=1}^{m}\frac{1}{2\sqrt{n}}$$
And after applying the telescoping property on the LHS, this becomes
$$\sqrt{m+1} - 1 < \sum_{n=1}^{m}\frac{1}{2\sqrt{n}}$$
But since $\sqrt{m} - 1 < \sqrt{m+1} - 1$, by transitivity we have
$$\begin{align*}\sqrt{m} - 1 &< \sum_{n=1}^{m}\frac{1}{2\sqrt{n}}\\
2\sqrt{m} - 2 &< \sum_{n=1}^{m}\frac{1}{\sqrt{n}}\end{align*}$$
Then the upper bound:
Proof (direct): From $(2)$ we have
$$\frac{1}{2\sqrt{n}}< \sqrt{n} - \sqrt{n-1}$$
By taking the sum of both sides we get
$$\sum_{n=1}^{m}\frac{1}{2\sqrt{n}}< \sum_{n=1}^{m}(\sqrt{n} - \sqrt{n-1})$$
If we apply the telescoping property on the RHS, the inequality becomes
$$\begin{align*}\sum_{n=1}^{m}\frac{1}{2\sqrt{n}} &< \sqrt{n}\\
\sum_{n=1}^{m}\frac{1}{\sqrt{n}} &< 2\sqrt{n}\end{align*}$$
Here I'm not sure how to get the $-1$ on the RHS that I'm lacking. I could have started the summation at $n=2$ I guess, but don't think it's correct and it doesn't seem to be a convincing way to complete the proof.
| For the lower bound, proof (direct): from $(1)$:
$$2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}}$$
By summing both sides:
$$2\sum_{n=1}^{m}(\sqrt{n+1} - \sqrt{n}) < \sum_{n=1}^{m}\frac{1}{\sqrt{n}}$$
After applying the telescoping property on the LHS, the inequality becomes
$$2\sqrt{m+1} - 2 < \sum_{n=1}^{m}\frac{1}{\sqrt{n}}$$
For the upper bound, proof (direct): from $(2)$:
$$\frac{1}{\sqrt{n}}< 2(\sqrt{n} - \sqrt{n-1})$$
By taking the sum of both sides from n=2:
$$\sum_{n=2}^{m}\frac{1}{\sqrt{n}}< \sum_{n=2}^{m}2(\sqrt{n} - \sqrt{n-1})$$
Add 1 on both sides:
$$1+\sum_{n=2}^{m}\frac{1}{\sqrt{n}}< 1+2(\sum_{n=2}^{m}(\sqrt{n} - \sqrt{n-1}))$$
$$\sum_{n=1}^{m}\frac{1}{\sqrt{n}}< 1+2(\sum_{n=2}^{m}(\sqrt{n} - \sqrt{n-1}))$$
Apply the telescoping property on the RHS, the inequality becomes
$$\begin{align*}\sum_{n=1}^{m}\frac{1}{\sqrt{n}} &< 1+2(\sqrt{m}-1)\\
\sum_{n=1}^{m}\frac{1}{\sqrt{n}} &< 2\sqrt{m}-1\end{align*}$$
So
$$2\sqrt{m+1} - 2 < \sum_{n=1}^{m}\frac{1}{\sqrt{n}} < 2\sqrt{m} - 1$$
This can be relaxed to:
$$2\sqrt{m}-2<2\sqrt{m+1}-2<\sum_{n=1}^{m}\frac{1}{\sqrt{n}}<2\sqrt{m}-1< 2\sqrt{m}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Precise limit definition Prove:
$$\displaystyle \lim_{x \to 2} \frac 1x = \frac 12.$$
We need to find a $\delta$ in terms of $\epsilon$.
Here is what I did so far:
\begin{align}
\left|\frac 1x-\frac 12 \right| &< \epsilon \\
-\epsilon < \frac 1x-\frac 12 &< \epsilon \\
-\epsilon + \frac 12< \frac 1x&< \epsilon + \frac 12 \\
\frac 2{1+2\epsilon} < x &< \frac 1{1-2\epsilon}
\end{align}
I am having trouble with this last step. Did I do this correctly?
| pick an $\epsilon$ positive and smaller than $1/2.$ let $x_l, x_r$ be defined by $\dfrac{1}{x_l} = \dfrac{1}{2} + \epsilon, \dfrac{1}{x_r} = \dfrac{1}{2} - \epsilon.$ computing $x_r- 2 = \dfrac{2}{1-2\epsilon} - 2=\dfrac{4\epsilon}{1-2\epsilon}$ and $2 - x_l = 2-\dfrac{2}{1+2\epsilon} =\dfrac{4\epsilon}{1+2\epsilon}$
choose $\delta = 4\epsilon$ so that $min \left\{2-x_l, 2-x_r \right\} = 2 - x_l < \delta.$ use the fact that $\dfrac{1}{x}$ is decreasing for $x > 0$ to conclude $$\lvert \dfrac{1}{x} - \dfrac{1}{2} \rvert < \epsilon \text{ for all } 2 -\delta < x < 2 + \delta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Integrate $\int\sqrt{1+9t^4}\:dt$ I need to find
\begin{align}
\int\sqrt{1+9t^4}\:dt.
\end{align}
What I have so far:
\begin{align}
\int\sqrt{1+9t^4}\:dt & =\int\sqrt{1+\left(3t^2\right)^2}\:dt,\tag{1}
\end{align}
now let $3t^2=\tan\left(\theta\right)\implies \displaystyle t=\sqrt{\frac{\tan\left(\theta\right)}{3}},\:\:dt=\frac{1}{2}\left(\frac{\tan\left(\theta\right)}{3}\right)^{-1/2}\left(\frac{\sec^{2}\left(\theta\right)}{3}\right)$, which gives me
\begin{align}
\int\left[1+\tan^2\left(\theta\right)\right]^{1/2}\frac{\sqrt{3}\sec^2\left(\theta\right)}{6\sqrt{\tan\left(\theta\right)}}\:d\theta\tag{2}&=\frac{\sqrt{3}}{6}\int\frac{\sec^3\left(\theta\right)}{\sqrt{\tan\left(\theta\right)}}\:d\theta\\
& = \frac{\sqrt{3}}{6}\int\frac{\sec^2\left(\theta\right)\sec\left(\theta\right)\:d\theta}{\sqrt{\tan\left(\theta\right)}}\tag{3},
\end{align}
now let $u=\tan\left(\theta\right),\:\:du=\sec^{2}\left(\theta\right)\:d\theta$, which gives us
\begin{align}
\frac{\sqrt{3}}{6}\int\frac{\sqrt{1+u^{\color{red}{2}}}\:du}{\sqrt{u}}\tag{4},
\end{align}
and where do I go from here? Or perhaps I'm just going in circles and haven't made any progress with this result.
| Substituting $u=t\sqrt3$ and $v=\dfrac1{1+u^4}$ we will be able to express this integral in terms of the
incomplete beta function of arguments $\dfrac14$ and $-\dfrac34$. Alternately, by expanding the integrand
into its binomial series, and switching the order of summation and integration, we can rewrite
the same integral in terms of the hypergeometric function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1145216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to build generating function when there is restriction on number of occurrences Let me ask my question in this way:
Suppose we are going to represent number $20$ as sum of $1$s and $2$s and $3$s, the coefficient of $x^{20}$ in this generating function is the answer :
$(1+x+x^2+x^3+...)(1+x^2+x^4+...)(1+x^3+x^6+...)$
Now if we put this restriction that the number of occurrences of $1$s must be more than number of occurrences of $2$s in the sequence of numbers that build $20$ then what generating function is going to be the answer. actually this problem can be generalized to many other cases.
| Summing the cases for the number of $1$s, we get
$$
\begin{align}
&\left(\vphantom{\frac12}\right.\overbrace{x\overbrace{\frac{1-x^2}{1-x^2}}^{1}}^{\text{one $1$ and $\lt$ one $2$}}+\overbrace{x^2\overbrace{\frac{1-x^4}{1-x^2}}^{1+x^2}}^{\text{two $1$s and $\lt$ two $2$s}}+\overbrace{x^3\overbrace{\frac{1-x^6}{1-x^2}}^{1+x^2+x^4}}^{\text{three $1$s and $\lt$ three $2$s}}+\dots\left.\vphantom{\frac12}\right)\overbrace{\overset{\vphantom{\overbrace{a^2}}}{\frac1{1-x^3}}}^{\text{any $3$s}}\\
&=\left(\frac{x}{1-x}-\frac{x^3}{1-x^3}\right)\frac1{(1-x^2)(1-x^3)}\\
&=\frac{x(1+x)}{(1-x^2)(1-x^3)^2}\\
&=\bbox[5px,border:2px solid #C00000]{\frac{x}{(1-x)(1-x^3)^2}}\tag{1}
\end{align}
$$
Now find the coefficient of $x^{20}$ in $(1)$.
We can compute the coefficients
$$
x\sum_{j=0}^\infty x^j\sum_{k=0}^\infty(k+1)x^{3k}
=\sum_{n=1}^\infty\frac12\left\lfloor\frac{n+2}3\right\rfloor\left\lfloor\frac{n+5}3\right\rfloor x^n\tag{2}
$$
since all $k\le\frac{n-1}3$ contributes $k+1$. That is,
$$
\begin{align}
\sum_{k=0}^{\left\lfloor\frac{n-1}3\right\rfloor}(k+1)
&=\frac12\left(\left\lfloor\frac{n-1}3\right\rfloor+1\right)\left(\left\lfloor\frac{n-1}3\right\rfloor+2\right)\\
&=\frac12\left\lfloor\frac{n+2}3\right\rfloor\left\lfloor\frac{n+5}3\right\rfloor
\end{align}
$$
For $n=20$, this gives $28$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Inverse Trig Functions Find $f(x)$ if $f'(x)=4/\sqrt{1-x^2}$ and $f(1/2)=1$
So far I have integrated $f'(x)$ and have found:
$$f(x) =y = 4\arcsin(x), x=4\sin(y)$$
$$1/2=4\sin(1)$$
$$1/2=4(\pi/2)$$
$$1/2=2\pi$$
So is $f(x)=1/2$ or $2\pi$?
Thanks
| If $f'(x) = \frac{4}{\sqrt{1-x^2}}$ then $f(x) = 4\arcsin(x) + c$, now if $f(1/2) = 1$ we have: $4\arcsin(1/2) + c = 1 \Rightarrow c = 1 - 4(\pi/6)$, therefore:
$f(x) = 4\arcsin(x) + 1 - 4(\pi/6)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that if $c\neq1$, the equation $ \frac{x}{x-a} + \frac{x}{x-b} = 1+c$ has exactly one real solution if $c^2 = - \frac {4ab}{(a-b)^2}$.
Show that if $c\neq1$, the equation
$$ \frac{x}{x-a} + \frac{x}{x-b} = 1+c$$ has exactly one real solution if $$c^2 = - \frac {4ab}{(a-b)^2}$$.
I know that a quadratic of the form $ax^2 +bx +c$ has only one real solution if $\sqrt{b^2-4ac} =0$, but I am not sure how to use this to solve the problem that has been set. I have also tried adding the two fractions and rearranging the result, but I have been unable to get anything resembling $ax^2 +bx +c$ .
| Since $x \neq a$ and $x \neq b$ you can multiply both sides of the equation by $(x - a)(x - b)$ arriving, after a bit of basic algebra, to
$$(1 - c)x^2 + c(a + b)x - ab(1 + c) = 0$$
Setting the condition $\Delta = 0$ will yield the desired result.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ \int_{-\infty}^{\infty} x^4 e^{-ax^2} dx$ What are some different methods to evaluate
$$ \int_{-\infty}^{\infty} x^4 e^{-ax^2} dx$$
for $a > 0$.
This integral arises in a number of contexts in Physics and was the original motivation for my asking. It also arises naturally in statistics as a higher moment of the normal distribution.
I have given a few methods of evaluation below. Anyone know of others?
| 1 Here's a relatively elegant method.
Notice that $\frac{\partial \ }{\partial a} e^{-a x^2} = - x^2 e^{-a x^2}$ and hence $\frac{\partial^2 \ }{\partial a^2} e^{-a x^2} = + x^4 e^{-a x^2}$
Thus, as the integrand is bounded and $C^\infty$ in both variables,
$$I = \int_{-\infty}^{\infty} x^4 e^{-ax^2} dx = \int_{-\infty}^{\infty} \frac{\partial^2 \ }{\partial a^2} e^{-ax^2} dx = \frac{d^2 \ }{da^2}\int_{-\infty}^{\infty} e^{-ax^2} dx $$
Since $\displaystyle \int_{-\infty}^{\infty} e^{-ax^2} dx = \frac{\sqrt \pi}{\sqrt a}$,
$$I = \frac{d^2 \ }{da^2} \frac{\sqrt \pi}{\sqrt a} = \frac{3\sqrt \pi}{4a^{5/2}}$$
2 Another method:
$$I^2 = \int_{-\infty}^{\infty} x^4 e^{-ax^2} dx \ \cdot \int_{-\infty}^{\infty} y^4 e^{-ay^2} dx = \int\int_{\mathbb R^2} (xy)^4 e^{-a(x^2 + y^2)} dx \ dy$$
Moving to polar coordinates,
$$I^2 = \int_0^{2\pi} \int_0^\infty r^8 \cos^4\theta\sin^4\theta e^{-ar^2} r \ dr \ d\theta = \int_0^\infty r^9e^{-ar^2} \ dr \ \cdot \ \int_0^{2\pi} \left(\frac{1}{2}\sin2\theta\right)^4 \ d\theta$$
With substitution $u = r^2$, the first integral is $\frac{4!}{2a^5}$. As $\sin^4 2\theta = \frac{1}{8} ( -4\cos4\theta + \cos 8\theta + 3)$, in the second integral the first two terms vanish over the domain of integration $[0,2\pi]$ and
$$I^2 = \frac{4!}{2a^5} \cdot \frac{1}{2^4} \frac{3}{8} 2\pi = \frac{9\pi}{16a^5}$$
Hence, as $I$ is positive,
$$I = \frac{3\sqrt \pi}{4a^{5/2}}$$
3 High school method:
Integrating by parts,
$$I = {-1 \over 2a} \int_{-\infty}^{\infty} x^3 (-2ax)e^{-ax^2} dx = {3\over 2a} \int_{-\infty}^{\infty} x^2 e^{-ax^2} dx $$
$$= {-3\over (2a)^2} \int_{-\infty}^{\infty} x (-2ax) e^{-ax^2} dx = {3 \over 4a^2} \int_{-\infty}^{\infty} e^{-ax^2} dx $$
and hence
$$I = {3 \over 4}{\sqrt\pi \over a^{5/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\frac{x^3-x^2-5x-3}{x^3-x^2-15x} \ge 0$ I've been trying to simplify the problem but I can't. I tried long division factoring and perfect cube but I can't still solve it. My calculator shows $x=3$ and $x=-1$ but wait how? this is for the numerator
| it seems you can factor $x^3-x^2 - 5x - 3 = (x+1)^2(x-3)$ and $x^3 - x^2-15x=x(x^2-x-15)=x(x -a)(x-b)$ where $a = \dfrac{1-\sqrt{61}}2, b = \dfrac{1+\sqrt{61}}2$
we need to solve $$y = \dfrac{(x+1)^2(x-3)}{(x-a)x(x-b)} \ge 0$$ you can draw a line and put the numbers $a, -1, 0, 3$ and $b$ and do a sign analysis or you can see that $y = 1+ \cdots$ for large $x$ and switches signs across $a, 0, 3$ and $b$ but at $x = 1$ it touches the $x$-axis from below. you will see that $$\dfrac{x^3-x^2 - 5x - 3}{x^3 - x^2 - 15x} \ge 0\text{ on }-\infty < x < a, 0 < x \le 3\text{ and }b < x < \infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Partial derivative problem on absolute value function Find the first and second order partial derivatives of
$f(x,y)=|2x^2-y|$.
I start with limit definition but not able to solve.
please help.
| We define $g$ as the function inside the absolute value. (i.e: $f(x,y) = |g(x,y)|$)
$$\frac{\partial f(x,y)}{\partial x} = \frac{g(x,y)\frac{\partial g(x,y)}{\partial x}}{|g(x,y)|}$$
Therefore, applying this to our function (i.e: $g(x,y) = 2x^2-y$) we get:
$$\frac{\partial |2x^2-y|}{\partial x} = \frac{(2x^2-y)(4x)}{|2x^2-y|},$$
$$\frac{\partial |2x^2-y|}{\partial y} = \frac{-2x^2+y}{|2x^2-y|},$$
for the first partial derivatives and
$$\frac{\partial\frac{(2x^2-y)(4x)}{|2x^2-y|}}{\partial x} = \frac{|2x^2-y|(24x^2-4y) - \frac{8x^3-4xy}{|2x^2-y|}}{|2x^2-y|^2},$$
$$\frac{\partial \frac{-2x^2+y}{|2x^2-y|}}{\partial y} = \frac{|2x^2-y| - \frac{(-2x^2+y)^2}{|2x^2-y|}}{|2x^2-y|^2},$$
for the second partial derivatives.
Finally, if we apply the definition of absolute value function to our results we get exactly what Statish Ramanathan said.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Putnam 1990 A1 Induction Help
A1. $(150,9,1,0,0,0,0,0,1,1,6,33)$
Let $$T_0=2,\quad T_1=3,\quad T_2=6,$$ and for $n\ge3$, $$T_n=(n+4)T_{n-1}-4nT_{n-2}+(4n-8)T_{n-3}.$$
The first few terms are $$2,3,6,14,40,152,784,5168,40576,363392.$$
Find, with proof, a formula for $T_n$ of the form $T_n=A_n+B_n$, where $(A_n)$ and $(B_n)$ are well-known sequences.
I found that:
$$T_n - n! = 2^n$$ and verified that $T_0 = 2^0 + 1 = 2$
Hence,
$$T_n = 2^n + n!$$
I need to prove that:
$$T_{n+1} = 2^{n+1} + (n+1)!$$
Which I can't do.
I went back and stated:
$$T_{n+1} = nT_{n} + T_n + 4T_n - 4nT_{n-1} - 4T_{n-1} + 4T_{n-2} + T_{n-2} - 8T_{n-2}$$
What to do next?
| Let's prove it by induction. At first we should establish the base case. $P(n_0) = T_0 = 0! + 2^0 = 1 + 1 = 2$ which is true. One can check the same with $T_1$ and $T_2$.
Now let's establish the strong induction hyphotesis which is as follows. $\forall n \ge 3$ and $ n \in \mathbb{Z} $ we have that $P(n ) = T_k = k! + 2^k $ for $ 1 \ge k \ge n. $
We shall prove that $T_{n+1} = (n+1) + 2^{n+1} $. To do so we're gonna perform the following algebra.
Using our strong hyphotesis and the given formula for $T_n$ set:
\begin{equation}
T_{n+1} = [(n+1)+4)(n! + 2^n)] -(4(n+1))[(n-1)! + 2^{n-1}] + (4(n-1))[(n-2)! + 2^{n-2}]
\end{equation}
After being multiplied out, the terms look like:
$$ T_{n+1} = 2^n + (n+1)! + 2^{n+2} + 2^n\cdot n + 4n! -4n! -4(n-1)! -2^{n+1}\cdot n - 2^{n+1} + 4(n-1)! + 2^n\cdot n - 2^n $$
As you can see, many terms will cancel out and in fact all the terms will (check by yourself), except by the following, giving us:
\begin{align*}
T_{n+1} &= (n+1)! + 2^{n+2} - 2^{n+1} \\ &= (n+1)! + 2^n(2^2 - 2) \\ &= (n+1)! + 2^n(4-2) \\ &= (n+1)! + 2^n(2) \\ &= (n+1)! + 2^{n+1} \blacksquare
\end{align*}
Quite elegantly done. QED.
| {
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"timestamp": "2023-03-29T00:00:00",
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Series $\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\cdots$ Find the sum of the series to infinity$$\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\cdots$$
Attempt-
I wrote the general term as $$\frac{\binom{2n}{n}}{2^{2n}\cdot (n+1)}$$
I don't know what to do next
| $$S = 2\left[\frac{1}{2\cdot 4}+\frac{1\cdot 3}{2\cdot 4\cdot 6}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}+\cdots\right]$$
Here $$T_{n}\bf{(n^{th}) term } = 2\cdot \left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6.........(2n)(2n+2)}\right]$$
$$ = 2\left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6...(2n)(2n+2)}((2n+2)-(2n+1))\right]$$
So $$T_{n} = 2\left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6.........(2n)}-\frac{1\cdot 3 \cdot 5......(2n+1)}{2\cdot 4 \cdot 6.........(2n)}\right]$$
So $$S_{\infty} = 2\sum^{\infty}_{r=1}\left[\frac{1\cdot 3 \cdot 5......(2r-1)}{2\cdot 4 \cdot 6.........(2r)}-\frac{1\cdot 3 \cdot 5......(2r+1)}{2\cdot 4 \cdot 6.........(2r)}\right] = 2\cdot \frac{1}{2} = 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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For which values of $a$, $b$ and $c$, if $a + b = c$, then $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$? I have a problem in my homework, which I have tried to solve, but I have just ideas, no real mathematical solutions. The problem is the following:
Suppose we have three real numbers $a$, $b$, and $c$ which satisfy the
equation:
$$a + b = c$$
Is it then also true that:
$$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$$
or not? Or is it only true for some particular choice of $a$, $b$, and
$c$, and which would that be?
My ideas:
*
*I noticed immediately that all $a$, $b$ and $c$ must be different from $0$, because otherwise we would have $\frac{1}{0}$ in the second equation, and that's not defined, as everybody knows.
*I tried to form a system of equations with the equations given in the specification of the problem:
$$\begin{cases} a + b = c \\ \frac{1}{a} + \frac{1}{b} = \frac{1}{c} \end{cases}$$
Since we have 3 variables ($a$, $b$ and $c$), I am not sure if this system of equations can bring me to some solution.
I have tried to replace $a + b$ in the second equation:
$$b(a + b) + a(a + b) = ab$$
$$ba + b^2 + a^2 + ba = ab$$
We can simplify to:
$$ba + b^2 + a^2 = 0$$
Now, I would not know how to proceed, and sincerely I don't know if my solution (ideas) is correct or not, or how far is it from the real solution.
My guess is that there's no values for $a$, $b$ or $c$ such that the 2 equations are valid.
| One more solution can't hurt, right?! We are trying to solve
\begin{align*} a+b=c && \frac{1}{a}+\frac{1}{b}=\frac{1}{c} && && (1) \end{align*}
where $a,b,c$ need to be nonzero numbers. As you pointed out, this is 3 unknowns and 2 equations. However, if we rearrange things a bit, we have the equivalent system
\begin{align*}
\frac{a}{c} + \frac{b}{c} = 1 && \frac{c}{a} + \frac{c}{b} =1.
\end{align*}
So you see, if we define $x=\frac{a}{c}$ and $y=\frac{b}{c}$ (two more nonzero numbers), then we get
\begin{align*}
x+y=1 && \frac{1}{x}+\frac{1}{y}=1. && && (2)
\end{align*}
Since the above system has two equations and two unknowns, it is more straightforward to convert it into a quadratic in one variable and see that Equation 2 has no solutions in the real numbers (although there are two complex solution pairs $(x,y)$).
It follows that there are no real solutions to Equation 1, although $(a,b,c) = (\lambda x, \lambda y, \lambda)$ will be a complex solution to Equation 1 whenever $(x,y)$ is a complex solution to Equation 2 and $\lambda$ is a nonzero complex number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1164973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq (b^2 + c^2)/2$? If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq \frac{b^2 + c^2}{2}$? I can only see how to get $a^2\leq \frac{b^2+c^2 + 2ab}{4}$.
| The original question asked us to show that if $a < b+c$ then $a^2 < b^2+c^2$ which is why you can see a lot of answers (including mine) that say it is false:
It isn't true in general! If $a=5,\, b=2$ and $c=4$ then:
$$5<2+4$$
But $$25 \not\lt 4 + 16$$
Edit: Answering the updated question:
If $a \le \frac{b+c}{2} $ then $$a^2 \le \left(\frac{b+c}{2}\right)^2.$$ And since $$\frac{b^2+c^2}{2} \ge \left(\frac{b+c}{2}\right)^2,$$ we get that $$a^2 \le \frac{b^2+c^2}{2}.$$
As required.
It is not immediately obvious how to prove the inequality used in the middle of that argument, $\frac{b^2+c^2}{2} \ge \left(\frac{b+c}{2}\right)^2$. But if you expand it out you find: $$\frac{b^2+c^2}{2} \ge \frac{b^2+c^2+2bc}{4},$$ $$2b^2+2c^2 \ge b^2+c^2+2bc,$$ $$b^2+c^2 \ge 2bc \qquad\mbox{for b, c > 0}.$$ Which can be seen to be true by rewriting $c$ as $b+\alpha$ for some $\alpha \in \mathbb{R}$ and working the algebra through:
$$b^2 + (b+\alpha)^2 \ge 2b(b+\alpha),$$
$$2b^2 + \alpha^2 + 2b\alpha \ge 2b^2 + 2b\alpha,$$
$$\alpha^2 \ge 0.$$ Which is true for all $\alpha$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Polynomial maximization: If $x^4+ax^3+3x^2+bx+1 \ge 0$, find the maximum value of $a^2+b^2$ If $x^4+ax^3+3x^2+bx+1 \ge 0$ for
all real $x$ where $a,b \in R$. Find the maximum value of $(a^2+b^2)$. I tried setting up inequalities in $a$ and $b$ but in the end had a hideous two variable expression whose maxima I had to calculate using Partial Derivatives. There must be a better way of doing it. Thanks!
| I'm still wondering about the "beautiful geometric solution," but there is another intuition that I think is useful.
The idea is that if $f(x) = x^4 + ax^3 + 3x^2 + bx + 1 \geq 0,$
then $f(x)$ can have only multiple roots (else its graph would cross the $x$ axis). A root of multiplicity $3$ is impossible (since it would imply another single root), and it is easy to show that a root of multiplicity $4$ also is impossible (since the terms $x^4$ and $1$ imply the root would have to occur at $x=\pm1,$ and this is inconsistent with the therm $3x^2$). Hence we have two or fewer roots of multiplicity $2.$
Briefly considering how $f(x)$ changes as we vary $a$ and $b,$
if $f(x) \geq 0$ for some pair of values of $a$ and $b$ of the same sign,
then we can reverse the sign of one of those coefficients while preserving
the property that $f(x) \geq 0.$ Hence whatever the maximum value of
$a^2 + b^2$ is, $a$ and $b$ have opposite signs.
The maximum value occurs when there is a double root on each side of the $y$ axis, since roots on only one side would enable one of the coefficients (either the positive one or the negative one) to have a larger magnitude.
In short, $f(x) = k(x - p)^2(x - q)^2$ for some positive number $k$ and some real numbers $p$ and $q.$
Multiplying this out, we have two ways to write $f(x)$ as a polynomial
in standard form,
\begin{align}
f(x) &= kx^4 - 2k(p + q)x^3 + (p^2 + 4pq + q^2)x^2 - 2kpq(p + q)x + kp^2q^2\\
&= x^4 + ax^3 + 3x^2 + bx + 1.
\end{align}
Comparing leading terms, we find that $k=1,$ and therefore the constant terms show that $p^2q^2 = 1,$ that is, $\lvert pq \rvert = 1$.
From the terms in $x$ and $x^3$ we see that $b = pqa$;
since $a$ and $b$ have opposite signs, $pq = -1$ and $q = -\frac1p.$
We can therefore equate the terms in $x^2$ as follows:
$$
p^2 - 4 + \left(\frac1p\right)^2 = 3.
$$
Setting $u=p^2,$ rearranging terms and multiplying by $u,$
we have $u^2 - 7u + 1 = 0,$ which has roots
$u = \frac12(7 \pm 3\sqrt5).$ We may recognize $\frac12(7 + 3\sqrt5)$ as the fourth power of the golden ratio.
Since $p = \pm\sqrt u,$ $p = \pm\frac12(3 \pm \sqrt5).$
Therefore $p$ and $q$ are either
$\frac12(3 + \sqrt5)$ and $-\frac12(3 - \sqrt5)$
or $-\frac12(3 + \sqrt5)$ and $\frac12(3 - \sqrt5)$ (in either order);
so $p + q = \pm\sqrt5,$ $a = \pm2\sqrt5,$ and $b = -a.$
Therefore $a^2 + b^2 = 40.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to evaluate $\int\frac{x}{\sqrt{x^2+x+1}}dx\;?$ How do you integrate,
$$\int \frac{x}{\sqrt{x^2+x+1}} dx $$
I am trying to use trig substitution, but I am having trouble finding a perfect square which works.
| Hint: $$\frac{x}{x^2+x+1} = \frac{x}{x^2+x+\frac{1}{4} + \frac{3}{4}} = \frac{x}{\left(x+\frac{1}{2}\right)^2 + \frac{3}{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\frac{a}c = \frac{a-b}{b-c}$ Suppose $\frac{1}a,\frac{1}b,\frac{1}c$ are three consecutive terms in an arithmetic sequence. Show that:
$$\frac{a}c = \frac{a-b}{b-c} $$
and that:
$$\frac{2ac}{a+c} = b$$
How would I prove this?
| $\dfrac{a}{c}=\dfrac{a-b}{b-c}\impliedby ab-ac=ac-bc\impliedby \color{red}{ab+bc=2ac}\impliedby\dfrac{1}{c}+\dfrac{1}{a}=\dfrac{2}{b}$
The second part follows from the colored part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$
| First, notice that for all $u$,
\begin{align*}(1-u)(1 + u + u^2) &= 1 + u + u^2 - (u - u^2 - u^3)\\
&= 1 + (u-u) + (u^2-u^2) - u^3\\
&= 1 - u^3.\end{align*}
Now, take $u = \frac{y}{x}$ and multiply by $x^3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove by mathematical induction that $\forall n \in \mathbb{N}: 20~|~4^{2n} + 4$ Prove by mathematical induction that $\forall n \in \mathbb{N}: 20~|~4^{2n} + 4$
Step 1: Show that the statement is true for n = 1:
$4^{2 \cdot 1} + 4 = 20$
Since $20~|~20$, the base case is completed.
Step 2: Show that if the statement is true for n = p, it is true for n = p + 1:
The general idea I had here was to try to get $4^{2(p+1)} + 4$ to look like the original statement, perhaps multiplied with a factor or a sum where each group of terms were divisible by 20.
$4^{2(p+1)} + 4 = 4^{2p+2} + 4$
By one of the exponential rules, we have:
$4^{2p+2} + 4 = 4^{2p} \cdot 4^{2} + 4$
At this point, we could use one of the 16 powers of 4 and together with the 4 term and show that this was divisible by 20, but there does not seem to be any clear reason why $20~|~4^{2p} \cdot 15$. The number $4^{2p}$ always have an even exponent and $4 \cdot 5 = 20$, but not sure how to proceed from here. Any suggestions?
| We assume that it is true for $n=p$, then $20|4^{2p} + 4$ so we can write $4^{2p} + 4$ as a product of 20 times an integer say, n, i.e. $4^{2p} + 4=20n$. Then,
$$4^2 (4^{2p} + 4)= 4^2 (20n) \\ 4^{2(p+1)} + 4^3 =4^2 (20n) \\ 4^{2(p+1)} + (64 -60) +60 =4^2 (20n) \\ 4^{2(p+1)} +4 = 20 \cdotp 16 \cdotp n -60 \\ 4^{2(p+1)} +4 =20(16n-3)$$
and since $16n-3$ is an integer, then $20| 4^{2(p+1)} +4$, which completes the proof by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find sequence for given generating function $\frac{ \frac{3x}{2} + \frac{3}{2}}{ 3 - x }$ I have generating function
$$A(x) = \frac{ \frac{3x}{2} + \frac{3}{2}}{ 3 - x }$$
and I need to find a sequence from it.
This was my approach:
$$ \frac{ \frac{3x}{2} + \frac{3}{2}}{ 3 - x } = \frac{-3}{2} + 2 \cdot \frac{1}{1 - \frac{x}{3}} $$
$ \frac{1}{1 - \frac{x}{3}} $ is generating function for $a_n = \frac{1}{3^n}$ so I though that the whole sequence $(c_n)_0^\infty$ would be:
$$c_n = \frac{-3}{2} + \frac{2}{3^n} $$
but according to wolfram it is wrong, where I am doing a mistake?
| $$\begin{align*}
\frac32\cdot\frac{x+1}{3-x}&=\frac12\cdot\frac{x+1}{1-\frac{x}3}\\
&=\frac12\left(x\sum_{n\ge 0}\left(\frac13\right)^nx^n+\sum_{n\ge 0}\left(\frac13\right)^nx^n\right)\\
&=\frac12\left(\sum_{n\ge 0}\left(\frac13\right)^nx^{n+1}+\sum_{n\ge 0}\left(\frac13\right)^nx^n\right)\\
&=\frac12\left(\sum_{n\ge 1}\left(\frac13\right)^{n-1}x^n+\sum_{n\ge 1}\left(\frac13\right)^nx^n+1\right)\\
&=\frac12\left(\sum_{n\ge 1}\left(\left(\frac13\right)^{n-1}+\left(\frac13\right)^n\right)x^n+1\right)\\
&=\frac12+\frac12\sum_{n\ge 1}\left(\frac43\right)\left(\frac13\right)^{n-1}x^n\\
&=\frac12+2\sum_{n\ge 1}\left(\frac13\right)^nx^n\;,
\end{align*}$$
so $a_0=\dfrac12$, and $a_n=\dfrac2{3^n}$ for $n\ge 1$.
| {
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Conditional probability and the binomial distribution
Let an unbiased die be cast at random seven independent times. Compute the conditional probability that each side appears at least once given that side 1 appears exactly twice.
We have $P[(\text{Each side appears} \geq 1) \mid (\text{Side 1 appears exactly exactly twice})] = \dfrac{P[(\text{Each side appears} \geq 1) \cap (\text{Side 1 appears exactly twice})]}{P[\text{Side 1 appears exactly twice}]}.$
For the numerator, if side 1 appears exactly twice, then the other 5 sides must appear exactly once. The probability of side 1 appearing exactly twice and the other 5 sides appearing exactly once is $(\frac{1}{6})^2(\frac{1}{6})^5$. There are $\binom{7}{2}$ ways of having side 1 appear exactly twice, and there are $5!$ ways the other 5 sides can appear exactly once.
Hence the numerator is $\binom{7}{2}\cdot5!\cdot(\frac{1}{6})^2(\frac{1}{6})^5.$
For the denominator, the 1 appearing exactly once has the probability $(\frac{1}{6})^2(\frac{5}{6})^5$. Considering that the 1's can be rolled in $\binom{7}{2}$ ways, the denominator is $\binom{7}{2}(\frac{1}{6})^2(\frac{5}{6})^5$.
When I divide the numerator by the denominator, I get $120\cdot\frac{1}{5^7} = \frac{24}{5^6},$ but the author has $\frac{24}{625} = \frac{24}{5^4}$, so somewhere I have two too many factors of 5.
| Your numerator and denominator are both correct. You can cancel $7\choose 2$ and $\left( \frac16 \right)^7$ top and bottom, leaving $\frac{5!}{5^5}=\frac{4!}{5^4}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof about exp function, inequality I need to proof the inequality, that $e^x > 1+x^2$ for positive $x$?
Thanks
Taylor doesn't work
Easy to proof for (0,1]. So one can assume x>1 in his proof
| Consider the Taylor expansion
$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots. $$
If $x \leq 2$ then $x \geq x^2/2$ and so $e^x > 1 + x + x^2/2 \geq 1 + x^2/2 + x^2/2 = 1 + x^2$.
If $x > 2$ then $x^3/6 > x^2/3$ and $x^4/24 > x^2/6$ and so $e^x > 1 + x^2/2 + x^3/6 + x^4/24 > 1 + x^2/2 + x^2/3 + x^4/6 = 1 + x^2$.
| {
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Need an explanation for binomial theorem question I understand the basic concept of binomial theorem, however once I begin looking at this question I can't seem to wrap my head around why x^0 is equivalent x^16 once you change the form of the question (This is off a short revision quiz). If anyone could give me a quick rundown on how to solve this question or even link me to a resource where I could figure it out, It would be much appreciated!
Thanks!
*Find the coefficient of $x^0$ in $(2x^2+5x^{-2})^8$.
Solution: Rewriting $(2x^2+4x^{-2})^8$ as $(x^{-2})^8(2x^4+5)^8$, we see that finding the coefficient of $x^0$ is equivalent to finding the coefficient of $x^{16}$ in $$(2x^4+5)^8=\sum_{i=0}^8\binom8{8-i}(2x^4)^{8-i}5^i\;.$$ Now $16=4(8-i)$, so $i=4$. Hence the coefficient is $$\binom842^45^4=\frac{8!}{4!4!}2^45^4=70\times 16\times 625\;.$$
| I’m assuming that you have no problem with the fact that $(2x^2+5x^{-2})^8=(x^{-2})^8(2x^4+5)^8$, but just in case: $2x^2+5x^{-2}=x^{-2}(2x^4+5)$.
We want the coefficient of $x^0$ in $(2x^2+5x^{-2})^8$; this is of course the same as the coefficient of $x^0$ in $(x^{-2})^8(2x^4+5)^8$. Note that $(x^{-2})^8(2x^4+5)^8=x^{-16}(2x^4+5)^8$, and suppose that we’ve multiplied out $(2x^4+5)^8$ and got some polynomial with terms like $ax^k$. When we finish the calculation by multiplying this polynomial by $x^{-16}$, each of its terms $ax^k$ will become $ax^{k-16}$. We want the term with exponent $0$ in the final product, which we’ll get when $k=16$. The coefficient won’t change when we multiply by the simple power $x^{-16}$, so the coefficient of $x^{16}$ in $(2x^4+5)^8$ will become the coefficient of $x^0$ in $x^{-16}(2x^4+5)^8$: the exponent will be reduced by $16$.
Knowing this, we can apply the binomial theorem to $(2x^4+5)^8$: it’s equal to
$$\sum_{i=0}^8\binom8{8-i}(2x^4)^{8-i}5^i=\sum_{i=0}^8\binom8{8-i}2^{8-i}x^{32-4i}5^i\;.$$
The exponent on $x$ is $16$ when $i=4$, and the coefficient is then
$$\binom8{8-4}2^{8-4}5^4=\binom842^45^4=10000\binom84=700,000\;.$$
| {
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When, how & who first gave this calculation of $\pi$ I came across this interesting method to calculate $\pi$. Why is it true and who first presented it?
To calculate $\pi$, multiply by $4$, the product of fractions formed by using, as the numerators. the sequence of primes $> 2$ ..i.e. $3,5,7,11,13,17$ etc and use as denominators the multiple of $4$ that is closest to the numerator and you will get as accurate value of $\pi$ that you like. In other words,
$$
\frac \pi 4 = \frac 3 4 \cdot \frac 5 4 \cdot \frac 7 8 \cdot \frac{11}{12} \cdot \frac{13}{12} \cdot \frac{17}{16}\cdots
$$
| This result is due to Euler, in a paper published in $1744$, entitled "Variae observationes circa series infinitas".
An English translation of the paper may be found here, and the quoted result on page $11$ in that link.
The proof follows from the series expansion for $\arctan$, giving:
$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\ldots$$
So that:
$$\frac{1}{3}\cdot\frac{\pi}{4}=\frac{1}{3}-\frac{1}{9}+\frac{1}{15}-\frac{1}{21}+\ldots$$
Adding the two together, we get rid of the $1/3, 1/9$ and all other multiples of $1/3$.
$$\frac{4}{3}\cdot\frac{\pi}{4} = \left(1+\frac{1}{3}\right)\frac{\pi}{4}
=1+\frac{1}{5}-\frac{1}{7}-\frac{1}{11}+\ldots$$
Repeat the same trick to get rid of the $1/5, 1/10, 1/15$ etc.:
$$\frac{4}{5}\cdot\frac{4}{3}\cdot\frac{\pi}{4} = \left(1-\frac{1}{5}\right)\frac{4}{3}\cdot\frac{\pi}{4}
=1-\frac{1}{7}-\frac{1}{11}+\frac{1}{13}+\ldots$$
If you continue this sequence, all compound factors get removed, and you are left with only with prime values in the denominator, and either one above or one below the prime in the numerator:
$$\cdots\frac{12}{11}\cdot\frac{8}{7}\cdot\frac{4}{5}\cdot\frac{4}{3}\cdot\frac{\pi}{4} = 1$$
Inverting gives the desired result:
$$\frac{\pi}{4} = \frac{3}{4}\cdot\frac{5}{4}\cdot\frac{7}{8}\cdot\frac{11}{12}\cdot\ldots$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Initial value problem for a linear system. Consider the linear system
$$
\frac{dY}{dt} =
\begin{pmatrix}
1 & -1 \\
1 & 3 \\
\end{pmatrix}
Y
$$
(a) Show that the function
$$
Y(t) =
\begin{pmatrix}
te^{2t} \\
-(t + 1)e^{2t}\\
\end{pmatrix}
$$
is a solution to the differential equation.
I verified this without trouble. The second part I am stuck on:
(b) Solve the initial-value problem
$$
\frac{dY}{dt} =
\begin{pmatrix}
1 & -1 \\
1 & 3 \\
\end{pmatrix}
Y
\text{, where }
Y(0) =
\begin{pmatrix}
0 \\
2 \\
\end{pmatrix}
$$
| Consider the differential system
\begin{align}
\frac{dY}{dt} =
\begin{pmatrix}
1 & -1 \\
1 & 3 \\
\end{pmatrix}
Y .
\end{align}
Let
\begin{align}
Y(t) = \begin{pmatrix} a(t) \\ b(t) \end{pmatrix}
\end{align}
such that
\begin{align}
\begin{pmatrix} a^{'} \\ b^{'} \end{pmatrix} = \begin{pmatrix} 1 & -1 \\
1 & 3 \\ \end{pmatrix} \cdot \begin{pmatrix} a \\ b \end{pmatrix}.
\end{align}
The set of differential equations is then given by
\begin{align}
a' &= a - b \\
b' &= a + 3b
\end{align}
of which $a$ and $b$ satisfy the second order equation $w'' - 4 w' +4 w = 0$. The solution of this second order equation is $w(t) = (A + B t ) e^{2t}$ and lead to the form of $Y(t)$ being
\begin{align}
Y(t) = \begin{pmatrix} A + B t \\ C + D t \end{pmatrix} \, e^{2t}.
\end{align}
Now the initial condition is
\begin{align}
Y(0) = \begin{pmatrix} 0 \\ 2 \end{pmatrix}
\end{align}
for which
\begin{align}
\begin{pmatrix} 0 \\ 2 \end{pmatrix} = \begin{pmatrix} A \\ C \end{pmatrix}
\end{align}
which yields the solution being
\begin{align}
Y(t) = \begin{pmatrix} B t \\ 2 + D t \end{pmatrix} \, e^{2t}.
\end{align}
Using this result to take its derivative and apply it to the differential equation it is seen that the result equations amount to $(B+D)t = -(B+2)$ and $(B+D)t = D-2$. In order for there to be solutions it is required that $B = -2$ and $D = 2$. The final solution becomes
\begin{align}
Y(t) = 2 \, e^{2t} \, \begin{pmatrix} - t \\ 1 + t \end{pmatrix}
\end{align}
| {
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Does $\sum_{n=1}^{\infty}\ln(n\sin(\frac{1}{n}))$ converge? I must determine whether the following series converges: $$\sum_{n=1}^{\infty}\ln\left(n\sin\left(\frac{1}{n}\right)\right)$$
I know that in general, I must use the limit comparison test, but I cannot find an expression to which I can compare it. For instance, I have tried the usual process:
For $n$ large, we have that $\lim_{n\to \infty}n\sin\frac1n=1$, and so, $\ln(1)=0$. This fails the divergence test, but it cannot be concluded automatically that the series is convergent either. How may I proceed here? Any help would be appreciated.
| Taylor series of $\sin x$ is
$$\sin x = x - \frac{x^3}{3!} + O(x^5)$$
Hence
$$\sin \frac{1}{n} = \frac{1}{n}- \frac{1}{6n^3} + O(\frac{1}{n^5})$$
So
$$n\sin \frac{1}{n} = 1- \frac{1}{6n^2} + O(\frac{1}{n^4})$$
Taylor series of $\ln$ is
$$\ln(1+x) = x + O(x^2)$$
For $x = -\frac{1}{6n^2} + O(\frac{1}{n^4})$, we get
$$\ln(1 - \frac{1}{6n^2} + O(\frac{1}{n^4})) = -\frac{1}{6n^2} + O(\frac{1}{n^4})$$
Then you'd get a sum of convergent series since all those sums have $p > 1$ according to the p-series test.
| {
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"timestamp": "2023-03-29T00:00:00",
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Having trouble solving $\int\frac{5x^2+3x+2}{x(x+1)^2}$ I've first transformed the integral to
$$\int\frac{5x^2+3x+2}{x(x^2+2x+1)}dx$$
Which gave me
$$\frac{A}{x}+\frac{Bx+C}{x^2+2x+1}$$
$$=\frac{A(x^2+2x+1)+Bx^2+Cx}{x(x^2+2x+1)}$$
$$\frac{5x^2+3x+2}{x(x^2+2x+1)}=\frac{(A+B)x^2+(2A+C)x+A)}{x(x^2+2x+1)}$$
So I've found the corresponding variables
$$A = 2$$
$$A+B = 5, B = 3$$
$$2A+C=3, C=-1$$
So the final integral is
$$2\int\frac{dx}{x}+3\int\frac{xdx}{x^2+2x+1}-\int\frac{dx}{x^2+2x+1}$$
$$=2ln(x) -ln(x^2+2x+1)+\frac{3}{x+1}+3ln(x+1)$$
However, the expected answer is
$$2ln(x)+3ln(x+1)+\frac{4}{x+1}$$
What is my error ?
| Hint: Notice that $x^2 + 2x + 1 = (x+1)^2$ then
$$\frac{5x^2 + 3x + 2}{x(x+1)^2} = \frac{A}{x} + \frac{B}{(x+1)} + \frac{C}{(x+1)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate when the degree of the numerator is higher than the denominator? I need to solve
$$\int\frac{x^5+4}{x^3+x^2}\,dx$$
I first tried to do a division, which gave me
$$\int\bigg[\frac{x^3}{x+1}+\frac{4}{x^2(x+1)}\bigg]\,dx$$
However, looking at the book, the integral is supposed to be transformed to
$$\int\bigg[x^2-x+1-\frac{4}{x}+\frac{4}{x^2}+\frac{3}{x+1}\bigg]\,dx$$
How am I suppose to end up with this integral ?
| You can also do this using series expansions. If you have a rational function $R(x) = \frac{p(x)}{q(x)}$ then you can consider the expansion around each zero of the denominator $q(x)$. If you keep only the singular terms and subtract all of those from $R(x)$ for each singularity, then you are left with a rational function that doesn't have any singularities left, which must therefore be a polynomial. That polynomial can, of course, be found using long division, but also by expanding around x = infinity (which is more efficient, this is equivalent to synthetic division).
The advantage of this method is that you are computing each term directly, so errors made in computing one term don't get propagated to other terms.
In this case, the expansion around x = 0 yields:
$$R(x) = (x^5+4)\frac{1}{x^2}\frac{1}{1+x} = (x^5+4)\frac{1}{x^2}(1-x +\mathcal{O}(x^2)) = \frac{4}{x^2} - \frac{4}{x}+\mathcal{O}(1)$$
The expansion around x = -1 yields
$$R(x) = (x^5+4)\frac{1}{x^2}\frac{1}{1+x} = \frac{3}{x+1} + \mathcal{O}(1)$$
The sum of all the singular terms of the expansions around all the singularities is thus:
$$S(x) = \frac{4}{x^2} - \frac{4}{x} + \frac{3}{x+1}$$
For large $x$ we can expand as follows:
$$R(x) = \frac{x^5+4}{x^3}\frac{1}{1+x^{-1}} = x^2 (1-\frac{1}{x} + \frac{1}{x^2})+\mathcal{O}(x^{-1}) = x^2 - x + 1+\mathcal{O}(x^{-1})$$
The expansion of $S(x)$ for large $x$ only contains negative powers of $x$, so we also have:
$$R(x) - S(x)= x^2 - x + 1+\mathcal{O}(x^{-1})$$
But we know that $R(x) - S(x)$ is a polynomial, therefore:
$$R(x) - S(x)= x^2 - x + 1$$
The partial fraction expansion is thus given by:
$$R(x) = x^2 - x + 1 + \frac{4}{x^2} - \frac{4}{x} + \frac{3}{x+1}$$
| {
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Indefinite integral of $ \int \frac{x^3}{\sqrt{x^2+1}} \,\text{d}x$. Can you please provide any sort of hint or suggestion on how to find the following indefinite integral?
$$\int\frac{x^3}{\sqrt{x^2+1}}\text{d}x$$
I tried substituting everything but it didn't work. I also tried trigonometric substitution but I couldn't find any valid trigonometric identity for $ \sqrt{\cos^2(x)+1} $ or $ \sqrt{\sin^2(x)+1} $.
| We have
$$\int \frac{x^3}{\sqrt{x^2 + 1}}\, dx = \int x \frac{(x^2 + 1) - 1}{\sqrt{x^2 + 1}}\, dx = \int x\sqrt{x^2 + 1}\, dx - \int \frac{x}{\sqrt{x^2 + 1}}\, dx.$$
Now use the $u$-substitution $u = x^2 + 1$ to get
$$\frac{1}{2}\int \sqrt{u}\, du - \frac{1}{2}\int \frac{du}{\sqrt{u}} =... $$
| {
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Simplification of an Equation with Recurrence Relations I'm reading through examples on this site.
In example 2_2, given the recurrence relation $A_n - 2A_{n-1} = 2n^2$, the guess for the particular solution is $A_n= Bn^2 + Cn + D$. Substituting that into the recurrence relation gets you
I just don't know how to continue this to get the answer above.
| As Robert Israel pointed out, the following is obtained.
Let $f_{n} = a n^{2} + b n + c$ in the difference equation $f_{n} = 2 f_{n-1} + 2 n^{2}$.
\begin{align}
a n^{2} + b n + c &= 2 a (n-1)^{2} + 2b (n-1) + 2c + 2 n^{2} \\
&= 2a n^{2} - 4b n + 2a + 2b n - 2b + 2 n^{2} \\
&= 2(a+1) n^{2} + (b-2a) n + (a-b+c).
\end{align}
This leads to
\begin{align}
(a + 2) n^{2} + (b-4a) n + (2a - 2b + c) = 0.
\end{align}
Equating coefficients leads to $a = -2$, $b= 4a = -8$, and $c = 2b -2a = -12$ for which
\begin{align}
f_{n} = -2( n^{2} + 4n + 6)
\end{align}
| {
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Proving Inequalities: AM-GM-HM, Cauchy-Schwarz/Rearrangement Inequalities.. Okay so I have two questions (I think they are pretty simple which was why I put them together), both relating to inequalities that are proving to be challenging. I have learned the AM-GM-HM Inequalities, the Rearrangement Inequality and the Cauchy-Schwarz Inequality.
A. Show that for positive reals $a, b, c$, such that $abc \le 1$
$$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$$
I have combed through this website and Google for solutions, hints anything but have for the most part come up empty handed. I have seen solutions very similar to it, but none have helped. I saw this exact solution:
It is easy to prove that
$$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge \frac{a + b + c}{\sqrt[3]{abc}}$$
and since $abc \le 1$then $\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$ as required.
It is not clear to me how they arrived at this in particular $\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge \frac{a + b + c}{\sqrt[3]{abc}}$. I feel like it is obvious, but I cant see it. So if someone could explain what inequality/trick was used, I would be grateful.
My attempt (which I think is completely wrong) is as follows.
$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$
by AM-GM inequality $\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge 3\sqrt[3]{\frac{a}{c}\cdot\frac{b}{a}\cdot\frac{c}{b}} = 3$
also by AM-GM $a+b+c \ge 3\sqrt[3]{abc}$
but $abc\le 1$
$\implies 3\sqrt[3]{abc} \le 3$
and thus
$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$
B. Let $a_1, a_2, ... , a_n$ be distinct positive integers. Prove that
$$\frac{a_1}{1^2} + \frac{a_2}{2^2} + ... + \frac{a_n}{n^2} \ge \frac{1}{1} + \frac{1}{2} +...+ \frac{1}{n}$$
A friend of mine decided to use the Rearrangement Inequality, but I don't really see that right off the bat. I tried to use Cauchy/Schwarz but I didn't get too far. If someone could give a hint or a nudge in the right direction as to which inequality I should use for this one I would appreciate it.
| For the first inequality, the trick is to apply the AM-GM inequality to each of the three terms in $${a\over{\sqrt[3]{abc}}} + {b\over{\sqrt[3]{abc}}}+ {c\over{\sqrt[3]{abc}}} .$$
(mouse over for spoiler)
$${a\over{\sqrt[3]{abc}}}=\sqrt[3]{a^3\over abc}=\sqrt[3]{\frac ab\cdot\frac ab\cdot\frac bc}$$
Hint for the second inequality: apply the Rearrangement Inequality to $a_1, a_2, \ldots,a_n$ and ${1\over1^2},{1\over2^2},\ldots,{1\over n^2}$. Once that's done, argue that the result follows.
| {
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Derivative of $f(x) = (x^2 +1)^3 (2x+5)^2$ I have function
$$f(x) = (x^2 +1)^3 (2x+5)^2$$
I need to find the derivative.
I believe that I need to use the product rule and chain rule.
Here's what I did.
$$f'(x) = (2x+5)^2[3(x^2+1)^2(2x)] + (x^2+1)^3[2(2x+5)2]
\\ = (2x+5)^2(6x(x^2+1)^2)+4(2x+5)(x^2+1)^3
\\ = 6x(2x+5)^2(x^2+1)^2+4(2x+5)(x^2+1)^3$$
That is what I have.
But the problem is, the answer book says that the answer is,
$$2(x^2+1)^2(2x+5)(8x^2+15x+2)$$
Since the answer book has been wrong few times, so I checked the answer with a calculator.
The calculator showed the same answer.
Then I used another calculator (which showed some steps), showed my answer.
The calculator showed,
Which answer is correct?? Or are they just equivalent answers??
If I did wrong, what was the problem??
Can it be factorized? How?
Thank you
| This is typical case where logarithmic differentiation makes life easy. $$f(x) = (x^2 +1)^3 \times(2x+5)^2$$ $$\log\big(f(x)\big)=\log(x^2+1)^3+\log(2x+5)^2=3\log(x^2+1)+2\log(2x+5)$$ Now, differentiate both sides $$\frac{f'(x)}{f(x)}=3\times \frac{2x}{x^2+1}+2\times\frac{2}{2x+5}$$ Now, reduce to same denominator $$\frac{f'(x)}{f(x)}= \frac{6x(2x+5)+4(x^2+1)}{(x^2+1)(2x+5)}=\frac{16 x^2+30 x+4}{(x^2+1)(2x+5)}=2\frac{8 x^2+15 x+2}{(x^2+1)(2x+5)}$$
Multiply now both sides by $f(x)$ and simplify to get $$f'=2(8 x^2+15 x+2)(x^2+1)^2(2x+5)$$ which is the answer from the book.
Do you find this way easier ?
| {
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Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1}$$
$$\cos (x)=\frac{1-u^2}{u^2+1}$$
$$dx=\frac{2du}{u^2+1}$$
$$=\int_{0}^{\infty }\frac{2(1-\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+1)}du$$
$$=\int_{0}^{\infty }\frac{2(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{(u+1)^2(u^2+1)}du $$
$$=2\int_{0}^{\infty }(\frac{2}{(u+1)^2}-\frac{1}{u^2+1})du $$
$$=-2\int_{0}^{\infty }\frac{1}{u^2+1}du+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du $$
$$\lim_{b\rightarrow \infty }\left | (-2\tan^{-1}(u)) \right |_{0}^{b}+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=(\lim_{b\rightarrow \infty}-2\tan^{-1}(b))+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=-\pi+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
Let $$s=u+1$$
$$ds=du$$
$$=-\pi+4\int_{1}^{\infty}\frac{1}{s^2}ds$$
$$=-\pi+\lim_{b\rightarrow \infty}\left | (-\frac{4}{s}) \right |_{1}^{b}$$
$$=-\pi+(\lim_{b\rightarrow \infty} -\frac{4}{b}) +4$$
$$=4-\pi$$
$$\approx 0.85841$$
| Substitute $x=\pi/2-2t$ so the integral becomes
$$
-2\int_{\pi/4}^{-\pi/4}\frac{1-\cos 2t}{1+\cos 2t}\,dt=
2\int_{-\pi/4}^{\pi/4}\frac{1-\cos^2t}{\cos^2t}\,dt
=2\Bigl[\tan t - t\Bigr]_{-\pi/4}^{\pi/4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 9,
"answer_id": 0
} |
Expand a function in Maclaurin's series. The function is given with:
$$\ln(5\cos^{3}(x))$$
Need to be expanded:
$$x^{4}$$
I have no idea what to do here and honestly, I don't really understand what a Maclaurin's series is. I know the definition but I don't understand the concept enough to be able to solve problems like this.
| $$\cos x=1-\frac{x^2}{2!}-\frac{x^4}{4!}+o(x^4)$$
and thus
$$5\cos^3(x)=5\left(1-\frac{x^2}{2!}+o(x^4)\right)^3=5-\frac{15x^2}{2!}+\frac{35}{8}x^4+o(x^4).$$
Moreover, if $x\longrightarrow 0$ you have that $5\cos^3(x)\longrightarrow 5$ and thus you have to develop $x\mapsto \ln(x)$ around $x=5$. That gives
$$\ln(x)=\ln(5)+\frac{1}{5}(x-5)-\frac{1}{25\cdot 2}(x-5)^2+o((x-5)^2).$$
Therefore
$$\ln(5\cos^3(x))=\ln(5)+\frac{1}{5}\left(-\frac{15x^2}{2!}+\frac{35}{8}x^4\right)-\frac{1}{50}\left(-\frac{15x^2}{2!}\right)^2+o(x^4)$$$$=\ln(5)-\frac{3}{2}x^2-\frac{1}{4}x^4+o(x^4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Lost on "Simple Computations" I have come across the follow assertion: $$\text{for } x,y,z >0, xyz = x+y+z+2$$ may be rewritten as $$\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=1$$
and that proving this is a matter of 'simple computations'. However, I can't seem to see the way about this.
I would say what I've tried so far, but every approach I've tried so far as petered out after a step or two.
| I would go reverse engineering for that one $$\frac{1}{x+1} + \frac{1}{1+y} + \frac{1}{1+z} = 1$$
First you add $\frac{1}{x+1} + \frac{1}{1+y}$ to get $$\frac{1+y+1+x}{(x+1)(y+1)}= \frac{2+x+y}{(x+1)(y+1)}$$ now we add this result to $\frac{1}{z+1}$ to get $$\frac{2+x+y}{(x+1)(y+1)} + \frac{1}{z+1} = \frac{(2+x+y)(1+z) + (x+1)(y+1)}{(x+1)(y+1)(z+1)} =1$$
now you have $(2+x+y)(1+z) + (x+1)(y+1) = (x+1)(y+1)(z+1) $ Now Subtract $(x+1)(y+1)$ from each side , you will end up with $$(2+x+y)(1+z) = (x+1)(y+1)(z+1)-(x+1)(y+1)$$ now we have $$(2+x+y)(1+z) = (x+1)(y+1)(z+1-1)$$ and so we have
$$(2+x+y)(1+z) = (x+1)(y+1)z$$ Now expand the LHS and the RHS to get $$2 + 2z + x + xz + y + yz = xyz + zx + zy + z$$. Now $xz$ cancels with $xz$ and $yz$ cancels with $yz$ and you end up with $$ 2 + 2z + x + y = xyz + z$$ Now final step , you subtract $z$ from each side to end up with $$2 + z + x + y = xyz$$ and you are done !!!!
If you don't like reverse engineering , you can reverse what I did, and you would be able to go from that to the fraction form. reverse reverse = direct :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solution congruence system $ x \equiv 11\pmod{36},\,x \equiv 7\pmod{40}, \,x \equiv 32\pmod{75}$ Have solution the following congruence system? $$\begin{align}
x & \equiv 11\pmod{36}\\
x & \equiv 7\pmod{40}\\
x & \equiv 32\pmod{75}
\end{align}$$
Point of Interest: This question requires some special handling due to the mixture of factors among the moduli. This is more than the run of the mill Chinese Remainder Theorem problem.
| Hint: It might be easier to break things down into unique and shared factors:
$$
\begin{align}
x&\equiv7\pmod{8}\\
x&\equiv2\pmod{9}\\
x&\equiv7\pmod{25}
\end{align}
$$
Once you have it in this form, you can then use the Extended Euclidean Algorithm to solve
$$
\begin{align}
a&\equiv1\pmod{8}\\
a&\equiv0\pmod{9\cdot25}\\
\end{align}
$$
$$
\begin{align}
b&\equiv1\pmod{9}\\
b&\equiv0\pmod{8\cdot25}\\
\end{align}
$$
$$
\begin{align}
c&\equiv1\pmod{25}\\
c&\equiv0\pmod{8\cdot9}\\
\end{align}
$$
and get
$$
x\equiv7a+2b+7c\pmod{8\cdot9\cdot25}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to calculate $\int \sqrt{x^2+6}\,dx$? How to calculate $\int \sqrt{x^2+6}\,dx$, by using Euler substitution and with to use of the formula : $\int u\,dv = vu - \int v\,du $.
note: what I mean by Euler substitution: is when we have a Integrand like $ \sqrt{x^2+1}$, then we can use the trick of substituting $t= x + \sqrt{x^2+1}$.
and that gives us: $dt=\frac{t}{\sqrt{x^2+1}} \, dx$, which can be helpful while solving the questions.
here I supposed that $u=\sqrt{x^2+6}$, and $du= \frac{x}{\sqrt{x^2+6}}\,dx$
then: $\int u\,dv = x\sqrt{x^2+6} - \int \frac{x^2}{\sqrt{x^2+6}}\,dx $, and then I got stuck with the latter integral. trying to substitute $t= x + \sqrt{x^2+6}$ in this case didn't really help. how can I do it $especially$ in this way? I know there are might be a lot of creative solutions, but I want some help continuing that direction.
| For the integral
\begin{align}
I = \int \sqrt{ x^{2} + 6 } \, dx
\end{align}
let $x = \sqrt{6} \sinh(t)$ to obtain $dx = \sqrt{6} \cosh(t) dt$ and
\begin{align}
I &= \int \sqrt{6} \, \sqrt{ 6 ( 1 + \sinh^{2}(t))} \, \cosh(t) \, dt \\
&= 6 \int \cosh^{2}(t) \, dt \\
&= 3 \int ( 1 + \cosh(2t)) \, dt \\
&= 3 \left[ t + \frac{\sinh(2t)}{2} \right] \\
&= 3 \left[ \sinh^{-1}\left( \frac{x}{\sqrt{6}} \right) + \frac{ x \, \sqrt{x^{2} + 6} }{ 6} \right]
\end{align}
Notes: Reading the question properly this solution would have been different. With that being said the presentation here is an alternate method of producing a solution to the proposed integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Converging sequence and its limit Show that the sequence $x_n$ given by
$$x_0=0, x_1=1 \ \ \text{ and} \ \ x_{n+1}=\sqrt{\frac{1}{4}x^{2}_{n}+\frac{3}{4}x^{2}_{n-1}}$$ for $n\in\mathbb N$
converges and find its limit.
My progress so far:
This is an intertwining sequence, and I am trying to show $I_{n+1}\subseteq I_n$ where interval $I_{n+1}=[x_{n}, x_{n-1}]$ by mathematical induction, but I failed to prove that case $n+1$ is true given that case $n$ is true. And I have no idea with the steps thereafter.
| The square root is making this difficult. Let's square both sides of your recursive expression:
$$x_{n+1}^2 = \frac{1}{4}x_n^2 + \frac{3}{4}x_{n-1}^2$$
let's consider the sequence $(y_n) = (x_n^2)$. Since each term in $(x_n)$ is nonnegative (by virtue of the square root), we have that $(x_n) = (\sqrt{y_n})$, so if $(y_n)$ converges to $y$, then $(x_n)$ will converge to $\sqrt{y}$.
We now have the linear recurrence relation
$$y_{n+1} = \frac{1}{4}y_n + \frac{3}{4}y_{n-1}$$
which we can rewrite in matrix form as
$$\left[\begin{array}{c} y_{n+1} \\ y_n\end{array}\right] = \left[\begin{array}{cc} \frac{1}{4} & \frac{3}{4} \\ 1 & 0\end{array}\right] \left[\begin{array}{c} y_{n} \\ y_{n-1}\end{array}\right]$$
so
$$\left[\begin{array}{c} y_{n+1} \\ y_n\end{array}\right] = \left[\begin{array}{cc} \frac{1}{4} & \frac{3}{4} \\ 1 & 0\end{array}\right]^n \left[\begin{array}{c} y_1 \\ y_0\end{array}\right]$$
If find that the eigenvectors of the matrix are $\left[\begin{array}{cc} 1 & 1\end{array}\right]^T$ and $\left[\begin{array}{cc} 1 & -\frac{4}{3}\end{array}\right]^T$, with eigenvalues of $1$ and $-\frac{3}{4}$, respectively. Thus, if we could write
$$\left[\begin{array}{c} y_1 \\ y_0\end{array}\right] = a \left[\begin{array}{c} 1 \\ 1\end{array}\right] + b \left[\begin{array}{c} 1 \\ -\frac{4}{3}\end{array}\right]$$
Then we would have
$$\left[\begin{array}{c} y_{n+1} \\ y_n\end{array}\right] = \left[\begin{array}{cc} \frac{1}{4} & \frac{3}{4}\\ 1 & 0\end{array}\right]^n \left[\begin{array}{c} y_1 \\ y_0\end{array}\right] = a(1)^n \left[\begin{array}{c} 1 \\ 1\end{array}\right] + b\left(-\frac{3}{4}\right)^n \left[\begin{array}{c} 1 \\ -\frac{4}{3}\end{array}\right]$$
as $n\to\infty$, the second term vanishes, so we would have $(y_n) \to a$, and thus $(x_n) \to \sqrt{a}$. I'll leave it to you to fill in the details.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Simplifying $\scriptsize\sqrt{2+\sqrt{2}} + \sqrt{2+\sqrt{2+\sqrt{2}}} + \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} + \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}$ The question is in the title: is there a simpler form or result for
$$\sqrt{2+\sqrt{2}} + \sqrt{2+\sqrt{2+\sqrt{2}}} + \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} + \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}\quad?$$
| $ \cos 2\theta =2\cos ^2\theta -1\Rightarrow \cos\theta=\sqrt{\dfrac{\cos 2\theta+1}{2}} $
Therefore $$\cos \dfrac{\pi}{8}=\sqrt{\dfrac{\frac{1}{\sqrt{2}}+1}{2}}=\sqrt{\dfrac{1+\sqrt{2}}{2\sqrt{2}}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}.$$
Hence $$\cos \dfrac{\pi}{16}=\sqrt{\dfrac{\dfrac{\sqrt{2+\sqrt{2}}}{2}+1}{2}}=\dfrac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}.$$
Then $$\cos \dfrac{\pi}{32}=\sqrt{\dfrac{\dfrac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}+1}{2}}=\dfrac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}.$$
Since $\sin^2 \dfrac{\pi}{32}+\cos^2 \dfrac{\pi}{32}=1$ we have $$\sin \dfrac{\pi}{32}=\dfrac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}.$$
So $$\sqrt{2+\sqrt{2}} + \sqrt{2+\sqrt{2+\sqrt{2}}} + \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} + \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}\\
=2\left(\cos \dfrac{\pi}{8}+\cos \dfrac{\pi}{16}+\cos \dfrac{\pi}{32}+\sin \dfrac{\pi}{32}\right)\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What will be the value of $\det(A^2+B^2)$? Let $A$ and $B$ be two $n\times n$ matrices such that $A \not=B, A^3 = B^3$ and $A^2 B = B^2 A$. Then what is the value of $\det \left( A^2 + B^2 \right)$?
Here I have done something.
$$A^3 = B^3 \implies(A-B)(A^2+B^2+AB)=0$$
Is my work right? Then how to do further? Well, I did mistakes here. But how can find the value of $\det(A^2+B^2)$?
| the following observation may help:
set $X=A^2+B^2$
then
$$
XA=A^3+B^2A = B^3 + A^2B = XB
$$
so
$$
X(A-B) = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplifying and evaluating $\cot 70^\circ+4\cos 70^\circ$ I have to simplify and evaluate this :
$$\cot 70^\circ+4\cos 70^\circ$$
On evaluating it, the answer comes out to be $1.732$, or $\sqrt 3$ .
I tried to get everything in $\sin$ and $\cos$, but it doesn't go any further. Any hints?
| Like Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $,
If $2\sin3x=-1$
$$\cot x+4\cos x=4\cos x-2\sin3x\cot x$$
$$=\dfrac{4\cos x\sin x-2\sin3x\cos x}{\sin x}$$
$$=\dfrac{2\sin2x-(\sin2x+\sin4x)}{\sin x}$$
$$=-2\cos3x\text{ as }\sin x\ne0$$
Here $x=70^\circ$
Generalization:
$$\sin3x=-\dfrac12\implies3x=180^\circ n+(-1)^n(-30^\circ)\text{ where $n$ is any integer}$$
$\implies x=60^\circ n+(-1)^n(-10^\circ)$ where $n\equiv-1,0,1\pmod3$
$n=-1\implies x=-50^\circ$
$n=0\implies x=-10^\circ$
$n=1\implies x=70^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Prove by induction on $n$ that when $x \gt 0$, $ (1+x)^n \ge 1+nx+\frac{n(n-1)}{2}x^2 \text{ for all positive integers } n. $ Here's the problem:
Prove by induction on $n$ that when $x \gt 0$
$$ (1+x)^n \ge 1+nx+\frac{n(n-1)}{2}x^2 \text{ for all positive integers } n. $$
So, clearly the base case is true. Here's how far I've gotten in the induction step:
$ (1+x)^{n+1} \ge 1+(n+1)x + \frac{(n+1)[(n+1)-1]}{2}x^2 $
$ (1+x)(1+x)^n \ge 1+nx+x+\frac{n^2+n}{2}x^2 $
$ (1+x)(1+x)^n \ge 1+[1+n+\frac{1}{2}(n^2+n)x]x $
...and that's pretty much it. I'm looking for an opportunity to use a triangle inequality, or maybe even Bernoulli's inequality, but nothing's popping out to me. Any hints would be appreciated. Thanks.
| To complete the induction
We have:
$$(1+x)(1+nx+\frac{n(n-1)}{2}x^2)=1+(n+1)x+nx^2+\frac{n(n-1)}{2}x^2+\frac{n(n-1)}{2}x^3$$
and using the inequality $x>0$ and $n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}x^2$ you can finish your work, indeed you get:
$$(1+x)(1+nx+\frac{n(n-1)}{2}x^2)\geq 1+(n+1)x+\frac{n(n+1)}{2}x^2$$
which gives the induction step!
Other method
$$(1+x)^n=\sum_{i=0}^n \dbinom k n x^k\geq \sum_{i=0}^3 \dbinom k n x^k=1+nx+\frac{n(n-1)}{2}x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find Matrix $A^{50}$? Find the matrix $A^{50}$ given
$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$
I was practicing some questions for my exam and I found questions of this form in a previous year's paper.
I don't know how to do such questions.
Please assist over this question.
Thank You
| And yet another approach: Splitting $A$ into a diagonal matrix $D$ and a pertubation matrix $P$.
We look at the first powers of $P$:
$$
P^0 =
\left(
\begin{matrix}
1 & 0\\
0 & 1
\end{matrix}
\right)
\quad
P^1 =
\left(
\begin{matrix}
0 & -1 \\
0 & 0
\end{matrix}
\right)
\quad
P^2 =
\left(
\begin{matrix}
0 & 0 \\
0 & 0
\end{matrix}
\right)
$$
and note that quadratic powers and higher of $P$ vanish.
Then
$$
DP = 2 P \quad PD = P
$$
Thus
\begin{align}
A^n
&= (D + P)^n \\
&= (D^2 + DP + PD + P^2) (D+P)^{n-2} \\
&= (D^2 + (1+2^1)P) (D+P)^{n-2} \\
&= (D^3 + (1+2^1)PD + D^2P + 3P^2)(D+P)^{n-3} \\
&= (D^3 + (1+2^1)P + 2^2 P)(D+P)^{n-3} \\
&= (D^3 + (1+2^1+2^2) P)(D+P)^{n-3} \\
&= D^n + (1+2^1+\cdots+2^{n-1})P \\
&= D^n + (2^n-1)P
\end{align}
This gives
$$
A^n
=
\left(
\begin{matrix}
2^n & 0 \\
0 & 1^n
\end{matrix}
\right)
+
\left(
\begin{matrix}
0 & 1-2^n \\
0 & 0
\end{matrix}
\right)
=
\left(
\begin{matrix}
2^n & 1-2^n \\
0 & 1
\end{matrix}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove that $\sqrt[3] 2 + \sqrt[3] 4$ is irrational? So while doing all sorts of proving and disproving statements regarding irrational numbers, I ran into this one and it quite stumped me:
Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.
I tried all the usual suspects like playing with $\sqrt[3]{2} + \sqrt[3]{4} = \frac{a}{b}$ for $a,b\in \mathbb{Z}$ , but got nowhere.
I also figured maybe I should play with it this way:
$2^\frac{1}{3} + 4^\frac{1}{3}=2^\frac{1}{3} + (2^2)^\frac{1}{3}=2^\frac{1}{3} + 2^\frac{2}{3}=2^\frac{1}{3} + 2^\frac{1}{3}\times 2^\frac{1}{3}=2^\frac{1}{3}(1+2^\frac{1}{3})$
But there I got stumped again, because while $1+2^\frac{1}{3}$ is irrational, nothing promises me that $2^\frac{1}{3} \times (1+2^\frac{1}{3})$ is irrational, and I feel like trying to go further down this road is moot.
So what am I missing (other than sleep and food)? What route should I take to
prove this?
Thanks in advance!
| An easy approach: If $p=\sqrt[3]{2}+\sqrt[3]{4}$ is rational:
$$p^2 = \sqrt[3]{4}+2\cdot 2+ 2\sqrt[3]{2} = p+4+\sqrt[3]{2}.$$
So $\sqrt[3]{2}=p^2-p-4$ would be rational.
An alternative, more general approach.
Claim: If $a,b$ are integers that are not perfect cubes, and $a\neq -b$, then $\sqrt[3]{a}+\sqrt[3]b$ is irrational.
Proof:
Assume $\sqrt[3]{a}+\sqrt[3]{b}$ is rational. Cube it and get:
$$(\sqrt[3]{a}+\sqrt[3]{b})^3 = a+ 3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) + b$$
Now, since $a,b$ are rational and $\sqrt[3]{a}+\sqrt[3]{b}$ is non-zero and rational, this means that $\sqrt[3]{ab}$ is rational.
Letting $p=\sqrt[3]{a}+\sqrt[3]{b}$ and $q=\sqrt[3]{ab}$, this means that
$$(x-\sqrt[3]{a})(x-\sqrt[3]{b}) = x^2-px+q$$ is a rational polynomial. It shares at least one root with $x^3-b$, But the GCD of these two polynomials has to be a rational polynomial, so the GCD cannot be linear (since it would be $x-\sqrt[3]b$, which is not a rational polynomial.)
This means that $x^2-px+q$ has to divide $x^3-b$. That means that $\sqrt[3]{a}$ is a root of $x^3-b$, which means that $a=b$. But there are no repeated roots of $x^3-b$, which yields a contradiction.
Corollary: If $a,b$ are rationals such that $a\neq -b$ and $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are irrational, then $\sqrt[3]{a}+\sqrt[3]{b}$ is irrational.
Proof: Rationalize the denominators and revert to the above theorem for integers.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "29",
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What skill do I lack to factor multivariate polynomials? Ok so I can factor easily regular quadratic polynomials, i.e. $5x^2+7x+9$ (I'm not sure whether that's prime, just made it up), and I was working on solving $y^2+(x^2+2x−2)y+(x^3−x^2−2x)$ by distributing the $y$ after the first parentheses and now I'm stuck. Am I just missing some skill in factoring multivariate polynomials? Don't tell me stuff about rings and fields because that won't help my situation. Thanks
| In this case, it's going to be taking advantage of what you know, to bootstrap yourself into new territory. You basically have
$$y^2 + By + C,$$
where $B = x^2 + 2x - 2$ and $C = x^3 - x^2 - 2x\ $ don't depend on $y$; it looks a little bit like a quadratic in $y$.
Using what we know about quadratics, we "know" it should factor as $(y + p)(y + q) = y^2 + (p + q)y + pq$, if it can be factored. Now, since you know how to factor quadratics with a single variable, that means you know we're looking for a pair of things (it would be a pair of numbers, if we had a good old quadratic with one variable), call them $p$ and $q$, such that $pq = C$ and $p + q = B$.
So, let's see how would could find solutions $p, q$ to $pq = C = x^3 - x^2 - 2x$.
We can factor $$C = x^3 - x^2 - 2x = x(x^2 - x - 2) = x(x - 2)(x + 1).$$
More specifically since we need a pair, let's see what two polynomials multiply to $C = x(x - 2)(x + 1)$:
\begin{align*} C &= x(x^2 - x - 2) \\
&=(-x)(-x^2 + x + 2)\\
&= (x^2 - 2x)(x + 1)\\
&= (-x^2 + 2x)(-x - 1)\\
&= (x^2 + x)(x - 2)\\
&= (-x^2 - x)(-x + 2)
\end{align*}
So, we have a few possibilities for $p$ and $q$, if we only require $pq = C$. Now let's see if any of them add up to $B = x^2 + 2x - 2$.
Lo and behold, $p = x^2 + x$ and $q = x - 2$ will work. We already know they multiply to $C$, and their sum is indeed $x^2 + 2x - 2 = B$.
Thus, your polynomial factors as
$$(y + p)(y + q) = (y + x^2 + x)(y + x - 2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
} |
$5^{th}$ power of any integer is of the form $11k$ or $11k +1$ or $11k -1$. $5^{th}$ power of any integer is of the form $11k$ or $11k +1$ or $11k -1$.
Let the integer be $x$. If $x$ has a factor $11$ then $x^5$ is of the form $11k$. Now we consider the case where $11 \not\mid x$ .
By Fermat's Theorem we know that $x^{10} \equiv 1(\mod 11)$. Thus $11 \mid x^{10} -1$ i.e. $11 \mid (x^{5} -1)(x^5 +1)$. Since $11$ is prime, $11 \mid (x^{5} -1)$ or $11 \mid (x^{5} +1)$. Thus $x^5$ is of the form $11k +1$ or $11k -1$.
Is the proof correct?
| Here is an alternative proof (though I find yours nicer):
*
*$x\equiv 0\pmod{11}\implies x^5\equiv 0^5\equiv11\cdot 0 \equiv 0\pmod{11}$
*$x\equiv 1\pmod{11}\implies x^5\equiv 1^5\equiv11\cdot 0+1\equiv+1\pmod{11}$
*$x\equiv 2\pmod{11}\implies x^5\equiv 2^5\equiv11\cdot 3-1\equiv-1\pmod{11}$
*$x\equiv 3\pmod{11}\implies x^5\equiv 3^5\equiv11\cdot 22+1\equiv+1\pmod{11}$
*$x\equiv 4\pmod{11}\implies x^5\equiv 4^5\equiv11\cdot 93+1\equiv+1\pmod{11}$
*$x\equiv 5\pmod{11}\implies x^5\equiv 5^5\equiv11\cdot 284+1\equiv+1\pmod{11}$
*$x\equiv 6\pmod{11}\implies x^5\equiv 6^5\equiv11\cdot 707-1\equiv-1\pmod{11}$
*$x\equiv 7\pmod{11}\implies x^5\equiv 7^5\equiv11\cdot1528-1\equiv-1\pmod{11}$
*$x\equiv 8\pmod{11}\implies x^5\equiv 8^5\equiv11\cdot2979-1\equiv-1\pmod{11}$
*$x\equiv 9\pmod{11}\implies x^5\equiv 9^5\equiv11\cdot5368+1\equiv+1\pmod{11}$
*$x\equiv10\pmod{11}\implies x^5\equiv10^5\equiv11\cdot9091-1\equiv-1\pmod{11}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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} |
Recursive equation with limit Find $\alpha, \beta, \gamma$ for recursive equation:
$$ \alpha a_{n+3}-3a_{n+1}+\beta a_n = 18n$$ $$a_0=0,a_1=\gamma, a_2=3 $$
$$\lim_{n\rightarrow\infty}\frac{3a_n+(-2)^{n}}{n^3}=3$$
Hey guys,
normally I know my way around recursive equations, but when it comes to exercise with limits like this one I don't know how to use the information from the limit condition. I looked in to answers and found out, that first you have to figure out that $a_n$ is formed like $n^3-\frac{1}{3}(-2)^n+ ...$
How can I come to this and move on from there?
My attempt was something like this:
$$\frac{3a_n+(-2)^{n}}{n^3}=3/*n^3$$
$$3a_n+(-2)^{n}=3n^3/-(-2)^{n}$$
$$3a_n=3n^3-(-2)^{n}/*\frac{1}{3}$$
$$a_n=n^3-\frac{1}{3}(-2)^{n}$$
but I don't know if this is correct and which parts are missing in $+...$ Please help
| First, convert the recursion to a linear recursion:
$$\alpha~a_{n+3}-3~a_{n+1}+\beta~ a_{n} = 18~n \tag{1}$$
$$\alpha~a_{n+4}-3~a_{n+2}+\beta~ a_{n+1} = 18~n +18 \tag{2}$$
$$\alpha~a_{n+5}-3~a_{n+3}+\beta~ a_{n+2} = 18~n + 36\tag{3}$$
Eliminate the nonlinear terms with linear algebra:
$$a_{n+5} = 2~a_{n+4} +
\left(\frac{3}{\alpha} - 1\right)~a_{n+3} +
\left(-\frac{\beta}{\alpha} - \frac{6}{\alpha}\right)~a_{n+2} +
\left(\frac{2\beta}{\alpha} + \frac{3}{\alpha}\right)~a_{n+1} -
\left(\frac{\beta}{\alpha} \right )~a_{n} \tag{4}$$
This is a linear recursion represented by the matrix:
$$\begin{bmatrix} a_{n+4} \\ a_{n+3} \\ a_{n+2} \\ a_{n+1} \\ a_{n+0} \end{bmatrix} =
\underbrace{\begin{bmatrix}
2 &
\frac{3}{\alpha} - 1 &
-\frac{\beta}{\alpha} - \frac{6}{\alpha} &
\frac{2\beta}{\alpha} + \frac{3}{\alpha} &
-\frac{\beta}{\alpha} \\
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
\end{bmatrix}^n}_M
\begin{bmatrix}a_{4} \\ a_{3} \\ a_{2} \\ a_{1} \\ a_{0}\end{bmatrix}
\tag{5}$$
For the recurrence to have a cubic $k^nn^3$ and an exponential $r^n$ term, then Jordan form of the matrix must contain subblocks:
$$\begin{bmatrix}
k & 1 & 0 & 0 \\
0 & k & 1 & 0 \\
0 & 0 & k & 1 \\
0 & 0 & 0 & k \\
\end{bmatrix} \text{ and } \begin{bmatrix} r \end{bmatrix}$$
Since the matrix is 5 by 5, with $k = 1$ and $r = -2$ the Jordan form can only be:
$${\rm Jordan}(M) =
\begin{bmatrix}
1 & 1 & 0 & 0 & 0\\
0 & 1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -2
\end{bmatrix}$$
So $M$ has an eigenvalue $1$ with multiplicity $4$ and eigenvalue $-2$ with multiplicity $1$. Expanding $\det (\lambda I - M) = 0$ you get:
$$\begin{cases}\lambda^5
- 2~\lambda^4
- \left(\frac 3{\alpha} - 1\right)~\lambda^3
+ \frac{\beta+6}{~\alpha}~\lambda^2
- \frac{2~\beta+3}{\alpha}\lambda + \frac \beta{\alpha} = 0 \\
(\lambda - 1)^4(\lambda + 2) = 0
\end{cases}$$
Solving gives $~\alpha=1$, $\beta=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Assistance with polynomial factorization Is it possible to factorize $x^6 + 4x + 8$?
If so what is the fully simplified form, and what would be the steps to get there?
| No, it,s not possible. Suppose $x^6 + 4x + 8=(x^n+f(x))(x^{6-n}+g(x))$, where $n\in\{1,2,3\}$ and $\deg f<n,\ \deg g<6-n$. Then since $x^6 + 4x + 8\equiv x^6(\mod2)$, we have $f\equiv 0(\mod4)$ and $g\equiv 0(\mod2)$. Write $f=2f^\prime$, $g=2g^\prime$. Since $x^6 + 4x + 8\equiv x^6(\mod4)$, we have $2x^ng^\prime+2x^{6-n}f^\prime\equiv 0(\mod4)$, or equivalently, $x^ng^\prime+x^{6-n}f^\prime\equiv 0(\mod2)$, which implies $g^\prime\equiv -x^{6-2n}f^\prime(\mod2)$. If $n=1,2$, the coefficients of $x^0$ and $x$ in $g$ is divisible by $4$, which implies the coefficient of $x$ in $(x^n+f(x))(x^{6-n}+g(x))$ is divisible by $8$, a contradiction; If $n=3$, then $g^\prime\equiv f^\prime(\mod2)$, thus $g\equiv f(\mod4)$, which also implies that the coefficient of $x$ in $(x^n+f(x))(x^{6-n}+g(x))$ is divisible by $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve a limit like this? $\lim_{x\to\infty} \left( 2^{\large\sin\left(\tfrac{x^2+5}{x+5}\right)}-2^{\large\sin(x-5)} \right)=0$ I've heard the Laurent Series is the best way of dealing with most limits but I'm not sure if Laurent Series can help with this exact one.
$$\lim_{x\to\infty} \left( 2^{\large\sin\left(\tfrac{x^2+5}{x+5}\right)}-2^{\large\sin(x-5)} \right)=0$$
I believe the reason why it approaches zero is because as you graph it the amplitude becomes significantly lower until up to x=10,000 it is near zero.
I tried to use Laurent series, but I can't use when a=0.
In my final step I substituted $\frac{1}{x}$ yet I couldn't isolate the variables to prevent the end result from being indeterminate for the Laurent series. It is also extremely difficult to use LHospital rule.
Is it possible to use the Laurent series at a=1 to solve the limit? Could you perhaps use the squeeze theorem?
| First, note that
$$
2^{\sin\left(\frac{x^2+5}{x+5}\right)}-2^{\sin(x-5)}=2^{\sin(x-5)}\left[2^{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}-1\right].
$$
Now, since $2^{\sin(x-5)}$ is bounded (it varies between $2$ and $1/2$), for the limit to be zero,
$$
\lim_{x\rightarrow\infty}2^{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}-1
$$
must be $0$.
In other words,
$$
\lim_{x\rightarrow\infty}2^{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}=1
$$
Note that
$$
\lim_{x\rightarrow\infty}2^{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}=2^{\lim_{x\rightarrow\infty}{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}}.
$$
Therefore, you must show that
$$
\lim_{x\rightarrow\infty}{\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)}=0.
$$
Observe that
$$
\lim_{x\rightarrow\infty}\frac{x^2+5}{x+5}-(x-5)=\lim_{x\rightarrow\infty}{\frac{x^2+5-(x^2-5)}{x-5}}=\lim_{x\rightarrow\infty}\frac{10}{x+5}.
$$
Therefore, the difference between $\frac{x^2+5}{x+5}$ and $x-5$ approaches zero. By the mean value theorem,
$$
\left(\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)\right)=\cos(c)\left|\frac{x^2+5}{x+5}-(x-5)\right|
$$
where $c$ is between $\frac{x^2+5}{x+5}$ and $x-5$ (since $\cos(x)$ is the derivative of $\sin(x)$). Since $\cos(x)$ is a bounded function,
$$
\lim_{x\rightarrow\infty}\left(\sin\left(\frac{x^2+5}{x+5}\right)-\sin(x-5)\right)=\lim_{x\rightarrow\infty}\cos(c)\left|\frac{x^2+5}{x+5}-(x-5)\right|\leq\lim_{x\rightarrow\infty}\left|\frac{x^2+5}{x+5}-(x-5)\right|=0.
$$
This completes the computation.
| {
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"url": "https://math.stackexchange.com/questions/1215867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sin^4\theta+\cos^4 \theta= 1-2\sin^2 \theta cos^2 \theta $ Is it possible to prove:
$$\sin^4\theta+\cos^4 \theta= 1-2\sin^2 \theta cos^2 \theta ?$$
| Another direct way would be to use $\cos^2\theta+\sin^2 \theta=1$ also in the forms $\sin^2 \theta = 1-\cos^2 \theta$ and $\cos^2 \theta = 1-\sin^2 \theta$ as follows:
$$\sin^4 \theta + \cos^4 \theta = \sin^2 \theta (1-\cos^2\theta)+\cos^2 \theta(1-\sin^2\theta)=1-2\sin^2\theta\cos^2\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Suppose that $a$ and $b$ belong to a field of order $8$ and that $a^2 + ab + b^2 =0$ then $a=0$ and $b=0$ .
Suppose that $a$ and $b$ belong to a field of order $8$ and $a^2 + ab + b^2 =0$. Then $a=0$ and $b=0$. Do the same when the field has order $2^n$ with $n$ odd?
If one of the term is zero, i.e. let $b=0$ then $a^2 =0 \implies a =0$. We also note that since field is of order $8$, $a^7 = 1 = b^7$.
But how to bring a contradiction if I consider any one to be not equal to $0$? I want to use only basics.
| First we have
$(a-b)(a^2+ab+b^2)=a^3-b^3=0$, or $a^3=b^3$.
Since $a^7=b^7=1,$
$a^7=a^4a^3=a^4b^3=1$, $b^7=b^3b^4=a^3b^4=1$, and $a^4b^3=a^3b^4$.
So $a^3b^3(a-b)=0$.
If $a^3=0$, then $a=0$, and then $b^2=0$, so $b=0$. Similarly it is true for $b^3=0$.
Finally, if $a^3\neq0$ and $b^3\neq0$, then $a=b$. So there is $3a^2=0$, and thus
$a=0$ and $b=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1218129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate $\sum_{m=1}^{N}\binom{m+k-1}{m}$. What would be a simplified formula for $\displaystyle \sum_{m=1}^{N}\binom{m+k-1}{m}$ for a given number $k$ and any number $N$?
| Use the following formula:
$$\displaystyle\binom{m+k-1}{m}=\binom{m+k}{m}-\binom{m+k-1}{m-1}$$
Proof:
\begin{align}
\binom{m+k}{m}-\binom{m+k-1}{m-1}&=\dfrac{(m+k)!}{m!\:k!}-\dfrac{(m+k-1)!}{(m-1)!\:k!}
\\
&=\dfrac{(m+k-1)!}{(m-1)!\:k!}\left(\dfrac{m+k}{m}-1\right)
\\
&=\dfrac{(m+k-1)!}{m!\:(k-1)!}
\\
&=\binom{m+k-1}{m}
\end{align}
So
\begin{align}
\sum \limits_{m=1}^{N}\binom{m+k-1}{m}&=\sum \limits_{m=1}^{N}(\binom{m+k}{m}-\binom{m+k-1}{m-1})
\\
&=\sum \limits_{m=1}^{N}\binom{m+k}{m}-\sum \limits_{m=1}^{N}\binom{m+k-1}{m-1}
\\
&=\sum \limits_{m=1}^{N}\binom{m+k}{m}-\sum \limits_{m=0}^{N-1}\binom{m+k}{m}
\\
&=\sum \limits_{m=1}^{N}\binom{m+k}{m}-\sum \limits_{m=1}^{N-1}\binom{m+k}{m}-1
\\
&=\binom{N+k}{N}-1
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to integrate $ \int \frac{x^2}{(x \sin(x)+\cos(x))^2} \mathrm{d}x$
Evaluate
$$\displaystyle \int \frac{x^2}{(x \sin(x)+\cos(x))^2} \mathrm{d}x$$
Can someone just tell me the necessary manipulations? Hints will be enough.
Can it be done by integration by parts?
| $\bf{My\; Solution::}$Given $$\displaystyle \int \frac{x^2}{\left(x \cdot \sin x+\cos x \right)^2}\mathrm{d}x$$
Now we can write $$\displaystyle x\cdot \sin x+1\cdot \cos x = \sqrt{1+x^2}\cdot \left(\sin x\cdot \frac{x}{\sqrt{1+x^2}}+\cos x\cdot \frac{1}{\sqrt{1+x^2}}\right)$$
$$\displaystyle =\sqrt{1+x^2}\cdot \cos \left(x-\phi\right).$$
Where $\displaystyle \sin \phi = \frac{x}{\sqrt{1+x^2}}$ and $\displaystyle \cos \phi = \frac{1}{\sqrt{1+x^2}}$ and $\displaystyle \tan \phi = x\Rightarrow \phi = \tan^{-1}(x).$
So our Integral now becomes $$\displaystyle = \int\frac{x^2}{1+x^2}\cdot \sec^2 (x-\phi)\mathrm{d}x$$.
Now Let $$(x-\tan^{-1}(x))=t\;,$$ Then $$\displaystyle \frac{x^2}{1+x^2}\mathrm{d}x = \mathrm{d}\phi.$$
So Integral Convert into $$\displaystyle \int \sec^2 t \ \mathrm{d}t = \tan t +\mathcal{C} = \tan \left(x-\tan^{-1}(x)\right)+\mathcal{C}$$.
$$\displaystyle = \frac{\tan x-x}{1+x\cdot \tan x}+\mathcal{C}=\frac{\sin x-x \cdot \cos x}{\cos x+x\cdot \sin x}+\mathcal{C}$$
So $$\displaystyle \int \frac{x^2}{(x\sin x+\cos x)^2}\mathrm{d}x = \frac{\sin x-x \cdot \cos x}{\cos x+x\cdot \sin x}+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Range of $f(x) =\frac {x -1}{x^2 -2x + 3} $? Is my solution for finding the range of
$$f(x) = \frac{x-1}{x^2 -2x + 3} $$
correct?
Since its Domain is $ \mathbb{R} $, so transforming this equation into $x$ in terms of $y$ , we get $$ yx^2 - (2y +1)x + (3y+1) = 0 $$
Now, since $ x \in \mathbb {R} $ , so its discriminant $\mathbb{D}$ must be $ \ge 0 $, so we have, $$ (2y + 1)^2 - 4y(3y + 1) \ge 0 $$
Solving it, we get
$$ y \in \left[-{1 \over {2 \sqrt2}}, {1 \over {2 \sqrt2}}\right] $$
Kindly solve my problem.
| Everything looks good, except you have a small mistake in your equation. Edit: There is no mistake anymore after OP edited his question
It should be $y x^2 - (2y + 1)x + 3y - 1 = 0$. This is solvable if and only if $$(2y + 1)^2 - 4y(3y - 1) = -8y^2 + 8y + 1 \ge 0.$$
The roots of $-8y^2 + 8y + 1 = 0$ are at
$$\frac{2 \pm \sqrt{6}}{4},$$
so we need to have $y \in \left[\frac{2 - \sqrt{6}}{4}, \frac{2 + \sqrt{6}}{4}\right]$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that matrices $\tiny\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix},\begin{pmatrix} 2& 0 \\ 1&2 \\ \end{pmatrix} $ are similar. Error in my method? Show that the matrix
$$\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix} $$ is similar to a triangular matrix of the form $$ \begin{pmatrix} \lambda& 0 \\ 1&\lambda \\ \end{pmatrix} $$ where $\lambda$ is an eigen value.
Attempt:
The eigenvalues of the matrix $\begin{pmatrix} 2&-1 \\ 0&2 \\
\end{pmatrix} $ are clearly $2,2$.
Hence, we need to show that the matrix $\begin{pmatrix} 2&-1 \\0&2 \\
\end{pmatrix} $ is similar to $\begin{pmatrix}2&0 \\1&2 \\\end{pmatrix} $
Now, we know that two matrices are similar if and only if they represent the same linear transformation.
But: for any two dimensional vector $(x ~~y)^T : $
$\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix} \begin{pmatrix}x \\y \\
\end{pmatrix} = \begin{pmatrix} 2x-y \\ 2y \\ \end{pmatrix} $
And
$\begin{pmatrix} 2&0 \\1&2 \\\end{pmatrix} \begin{pmatrix}x \\ y \\\end{pmatrix} = \begin{pmatrix} 2x \\x+2y \\\end{pmatrix} $
Clearly, $\begin{pmatrix} 2x \\x+2y \\\end{pmatrix} $ and $\begin{pmatrix}
2x-y \\2y \\ \end{pmatrix} $ don't represent the same linear transformation.
So, how can these matrices be similar?
What could be the fault in my reasoning?
Thank you very much for your help in this regard.
| Call $A$ the first matrix, written with respect to a basis $e_{1}, e_{2}$ and $B$ the second, written with respect to a basis $f_{1}, f_{2}$.
Note that $A$ fixes $e_{1}$, and $B$ fixes $f_{2}$. Moreover, $A e_{2} = 2 e_{2} - e_{1}$. Can you find a number $a$ so that for the vector $f_{1} + a f_{2}$ one has
$$
B (f_{1} + a f_{2}) = 2 (f_{1} + a f_{2}) - f_{2}?
$$
If you can do that, the matrix of $B$ with respect to the basis $f_{2}, f_{1} + a f_{2}$ will be $A$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $(M,+)$ is isomorphic to $(\mathbb{C},+)$
Let, $M=\{ \begin{pmatrix}a&-b\\b&a\end{pmatrix} :a,b\in \mathbb{R}\}$, show $(H,+) $ is isomorphic as a binary structure to $(\mathbb{C},+)$
Define $f : M\rightarrow \mathbb{C} $ by $f\begin{pmatrix}a&-b\\b&a\end{pmatrix} = a+bi$
Let $a,b,c,d\in \mathbb{R}$
$1-1$:
Suppose $f\begin{pmatrix}a&-b\\b&a\end{pmatrix}=f\begin{pmatrix}c&-d\\d&c\end{pmatrix}$, then $a+bi =c+di$, thus $a=c$ and $b=d$, so $f$ is one to one.
Onto:
Let $a+bi\in \mathbb{C}$ , then $\begin{pmatrix}a&-b\\b&a\end{pmatrix}\in M$, so $f\begin{pmatrix}a&-b\\b&a\end{pmatrix}=a+bi$, thus $f$ is onto.
Homomorphic:
$\begin{align*} &f(\begin{pmatrix}a&-b\\b&a\end{pmatrix} + \begin{pmatrix}c&-d\\d&c\end{pmatrix})\\
=&f(\begin{pmatrix}a+c&-(b+d)\\b+d&a+c\end{pmatrix})\\
= &(a+c)+(b+d)i\\
= &f(\begin{pmatrix}a&-b\\b&a\end{pmatrix}) +f(\begin{pmatrix}c&-d\\d&c\end{pmatrix})\end{align*}$
I don't think I show $f: M\rightarrow \mathbb{C} $ is $1-1$ and onto correctly.
can any give me a hit to show $f: M\rightarrow \mathbb{C}$ is $1-1 $ and onto? thanks!
| Responding to the request from @Julian and to move this question to answered question list, I will try to sum up what has been mentioned in the comments.
Yes, your answer is correct.
| {
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A real analysis proof question (related to sin(1/x)) The problem statement is
Let
$$f(x) = \left\{\begin{array}{cc}
x^4 \left(2 + \sin \frac 1 x\right) & x \ne 0 \\
0 & x = 0
\end{array}\right.
$$
(a)Prove that $f$ is differentiable on $\mathbb{R}$
(b)Prove that $f$ has an absolute minimum at $x=0$
(c) Prove that $f'$ takes both positive and negative values in every neighborhood of $0$.
This first two parts of the problem are pretty straightforward. The only problem I encountered was in the last part. I was not sure how to prove it. I know both $\sin(1/x)$ and $\cos(1/x)$oscillate near zero. For any interval around zero they are gonna take positive and negative values since they oscillates. But this was not consider as a "proof". Is there a way to do that more rigorously? Can I use intermediate value property?
BTW, $$f'(x) = 4x^3\left(2+\sin \frac 1 x\right)+x^2\cos \left(\frac 1 x\right)$$ when $x\neq 0$.
| Since away from $0$ we have
$$f(x) = x^4\left(2 + \sin \frac 1 x\right)$$
we have
\begin{align*}
f'(x) &= 4x^3 \left(2 + \sin \frac 1 x\right) + x^4 \left(\cos \frac 1 x\right) \left(-\frac 1 {x^2}\right) \\
&= 8x^3 + 4x^3 \sin \frac 1 x - x^2 \cos \frac 1 x \\
&= x^2 \left(8x + 4x \sin \frac 1 x - \cos \frac 1 x\right)
\end{align*}
(Note there's a sign error in the original post, but it doesn't actually matter). Now notice that the terms with $x$ tend to zero at the origin while the cosine term does not: The cosine term is dominant here. To be explicit, choose $x = 1 / (n \pi)$ with $n > 8$. The first two terms can be controlled as
$$\left|8x + 4x \sin \frac 1 x\right| \le 12 |x| < \frac 1 2$$
On the other hand, $\cos 1/x = \cos (n\pi) = \pm 1$ according to the parity of $n$. Hence, the derivative can be estimated either as
$$f'(x) > \frac 1 {(n\pi)^2} \cdot \left(1 - \frac 1 2\right) = \frac 1 {2 (n \pi)^2}$$
if $n$ is odd or
$$f'(x) < \frac 1 {(n \pi)^2} \cdot \left(-1 + \frac 1 2\right) = - \frac 1 {2(n\pi)^2}$$
if $n$ is even.
| {
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"url": "https://math.stackexchange.com/questions/1233946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\int_0^1 \frac{dx}{f^2(x)+1} \le \frac{ \pi}{4}$ Let $f:[0,1] \to \mathbb{R}$ be a differentiable function, for which $f'(x) \ge 1 , \forall x\in [0,1]$, and $f(1)=1$. Prove that: $$\int_0^1 \frac{dx}{f^2(x)+1} \le \frac{ \pi}{4}$$
From the hyphotesis, we deduce that $f(0) \le 0$. This doesn't help very much, because if we write $$\int_0^1 \frac{dx}{f^2(x)+1} \le \int_0^1 \frac{f'(x)}{f^2(x)+1} dx =\arctan(f(x))|_0^1=\frac{ \pi }{4}-\arctan(f(0)) \ge \frac{ \pi }{4}$$
Another attempt is: We notice that $ \frac{ \pi }{4} = \int_0^1 \frac{dx}{x^2+1}$, so we only need to prove that $\int_0^1 \frac{dx}{f^2(x)+1} \le \int_0^1 \frac{dx}{x^2+1}$. This can be written as:
$$ 0 \le \int_0^1 \frac{(f(x)-x)(f(x)+x)}{(x^2+1)(f^2(x)+1)} dx $$
whichi, I think, isn't very easy to prove.
Also, I tried to use that the function $x \to f(x)-x$ is increasing (in fact, I used it when I proved that $f(0) \le 0$), but nothing.
| As an addition to the answer above:
It is sufficient to look at linear function $f(x) = \frac{x-a}{1-a}$ with sufficient small $a \in [0, 1]$
$$ \int^1_0{\frac{dx}{1 + f^2(x)}} = \int^1_0{\frac{dx}{1 + (\frac{x-a}{1-a})^2}} = I $$
Then $ u = \frac{x-a}{1-a}$, $du = \frac{dx}{1-a}$
$$I = (1-a)\int^1_{\frac{-a}{1-a}}{\frac{du}{1 + u^2}} = \frac{\pi}{4} - a\frac{\pi}{4} + (1-a)\arctan(\frac{a}{1-a})$$
Now for sufficiently small $a$ (using first two terms in arctan series):$$I > \frac{\pi}{4} + a(1-\frac{\pi}{4} - \frac{a^2}{3(1-a)^2}) > \frac{\pi}{4}$$
Where "sufficiently small" means: $$a < \frac{\gamma}{1 + \gamma}, \gamma = \sqrt{3(1 - \frac{\pi}{4})}$$
| {
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"url": "https://math.stackexchange.com/questions/1236073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Something wrong at $\int \frac{x^2}{x^2+2x+1}dx$ I have to calculate $$\int \frac{x^2}{x^2+2x+1}dx$$ and I obtain: $$\int \frac{x^2}{x^2+2x+1}dx=\frac{-x^2}{x+1}+2\left(x-\log\left(x+1\right)\right)$$
but I verify on wolfram and this is equal with: $$x-\frac{1}{x+1}-2\log\left(x+1\right)$$ where did I go wrong?
P.S: Here is all steps: $$\int \:\frac{x^2}{x^2+2x+1}dx=-\frac{x^2}{x+1}+\int \:\frac{2x}{x+1}dx=-\frac{x^2}{x+1}+2\left(x-\log\left(x+1\right)\right)$$ I am sure that is not wrong, but the form on wolfram seems easily then it.
| Hint: write $$\frac{x^{2}}{\left(x+1\right)^{2}}=1+\frac{1}{\left(x+1\right)^{2}}-\frac{2}{x+1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving an Inequality (terms won't cancel out) Problem: If $x$ and $y$ are real numbers such that $y \geq 0$ and $y(y+1) \leq (x+1)^2$, prove that $y(y-1) \leq x^2$.
This is what I tried:
\begin{align}
y(y+1) \leq (x+1)^2 &\implies y^2 + y \leq x^2 + 2x + 1 \\
&\implies x^2 \geq y^2 + y - 2x - 1 \\
&\implies x^2 \geq y^2 - y - 2x - 1 \tag{because $y \geq 0$}\\
&\implies x^2 \geq y(y-1) -2x - 1
\end{align}
I think I'm approaching this wrong, because if I start with the inequality $y(y+1) \leq (x+1)^2$, then the $2x + 1$ that results from the expansion of the RHS will never cancel out.
So, I tried this instead:
\begin{align}
y(y+1) \leq (x+1)^2 &\implies y^2 + y \leq x^2 + 2x + 1 \\
&\implies x^2 \geq y^2 + y - 2x - 1 \\
&\implies x^2 \geq y^2 - y - 2x - 1 \tag{same as before}
\end{align}
Then, I supposed that $y(y-1) \leq x^2$ is true. Consequently, because $y^2 - y - 2x - 1 \leq x^2$ (the last line), it is sufficient to show that $y(y-1) \leq y^2 - y - 2x - 1$. Again, this doesn't work because $-2x - 1$ doesn't cancel out.
Now I'm very much stuck and don't know how to proceed. If anyone can give me a hint, it'll be very much appreciated. Thanks!
| Consider two cases: $2x+1 \leq 2y$, and $2x+1 > 2y$. Note that we can assume $y > 1$ since if $y < 1$ the inequality is true trivially.
Specifically, if $2x+1 \leq 2y \to y(y-1) = y(y+1) - 2y \leq (x+1)^2 - 2y = x^2 + (2x+1-2y) \leq x^2 + 0 = x^2$, and if $2x+1 > 2y \to x > \dfrac{2y-1}{2} \to x^2 > \dfrac{(2y-1)^2}{4} = \dfrac{4y^2-4y+1}{4} = y(y-1) + \dfrac{1}{4} > y(y-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this integral by a simple way? I'm given $$\int \frac{x^3}{\sqrt{x^4+x^2+1}}dx$$
My attempt,
Let $u=x^2$, $du=2xdx$
$$=\frac{1}{2}\int \frac{u}{\sqrt{u^2+u+1}}du = \frac{1}{2}\int \frac{u}{\sqrt{(u+\frac{1}{2})^2+\frac{3}{4}}}du$$
Let $s=u+\frac{1}{2}$, $ds=du$
$$=\frac{1}{2}\int\frac{s-\frac{1}{2}}{\sqrt{s^2+\frac{3}{4}}} ds$$
Let $s=\frac{\sqrt{3}}{2}\tan p$, $ds=\frac{\sqrt{3}}{2}\sec^2 p\,dp$
So, $$\sqrt{s^2+\frac{3}{4}}=\sqrt{\frac{3\tan^2 p}{4}+\frac{3}{4}} =\frac{\sqrt{3}}{2}\sec p$$
and
$$p=\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)$$
$$=\frac{\sqrt{3}}{4}\int \frac{2\left(\frac{\sqrt{3}}{2}\tan p-\frac{1}{2}\right)\sec p}{\sqrt{3}}dp$$
$$=\frac{1}{2}\int \left(\frac{\sqrt{3}}{2}\tan p-\frac{1}{2}\right)\sec p\,dp$$
$$=\frac{\sqrt{3}}{4}\int \tan p\sec p\,dp-\frac{1}{4}\int \sec p\,dp$$
Let $w=\sec p$, $dw=\tan p\sec p\,dp$
$$=\frac{\sqrt{3}}{4}\int 1 dw-\frac{1}{4}\int \sec p\,dp$$
$$=\frac{\sqrt{3}w}{4}-\frac{1}{4}\int \sec p\,dp$$
$$=\frac{\sqrt{3}w}{4}-\frac{1}{4}\ln (\tan p+\sec p)+c$$
$$=\frac{1}{4}\sqrt{3}\sec p-\frac{1}{4}\ln (\tan p+\sec p)+c$$
$$=\frac{1}{4}\sqrt{3}\sec \left(\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)\right)-\frac{1}{4}\ln \left[\tan\left(\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)\right)+\sec \left(\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)\right)\right]+c$$
$$=\frac{1}{4}\sqrt{4s^2+3}+\frac{1}{8}\left[\ln 3 -2\ln \left(\sqrt{4s^2+3}+2s\right)\right]+c$$
$$=\frac{1}{2}\sqrt{u^2+u+1}+\frac{1}{8}\left[\ln 3-2\ln \left(2\sqrt{u^2+u+1}+2u+1\right)\right]+c$$
$$=\frac{1}{2}\sqrt{x^4+x^2+1}+\frac{1}{8}\left[\ln 3-2\ln \left(2x^2+2\sqrt{x^4+x^2+1}+1\right)\right]+c$$
$$=\frac{1}{8}\left[4\sqrt{x^4+x^2+1}-2\ln \left(2x^2+2\sqrt{x^4+x^2+1}+1\right)+\ln 3\right]+c$$
| Some ideas:
$$\int\frac{x^3}{\sqrt{x^4+x^2+1}}dx=\frac14\int\frac{\overbrace{4x^3+2x}^{=(x^4+x^2+1)'}}{\sqrt{x^4+x^2+1}}dx-\frac14\overbrace{\int\frac{2xdx}{\sqrt{x^4+x^2+1}}}^{u:=x^2\implies du=2xdx}=$$
$$\frac12\sqrt{x^4+x^2+1}-\frac14\int\frac{du}{\sqrt{u^2+u+1}}=\frac12\sqrt{x^4+x^2+1}-\frac14\int\frac{du}{\sqrt{\left(u+\frac12\right)^2+\frac34}}=$$
$$=\frac12\sqrt{x^4+x^2+1}-\frac1{2\sqrt3}\int\frac{du}{\sqrt{\left(\frac{\sqrt3}2u+\frac{\sqrt3}4\right)^2+1}}=$$
$$=\frac12\sqrt{x^4+x^2+1}-\frac13\int\frac{\frac{\sqrt3}2du}{\sqrt{\left(\frac{\sqrt3}2u+\frac{\sqrt3}4\right)^2+1}}=$$
Finally, change variables: $\;\sinh t=\dfrac{\sqrt3}2u+\dfrac{\sqrt3}4\;,\;\;\cosh t\,dt=\dfrac{\sqrt3}2du\;$
and get a very, very simple last integral above...and then go back to the original variable if you wish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Simplify $\frac{(\cos \frac{π}{7}-i\sin\frac{π}{7})^3}{(\cos\frac{π}{7}+i\sin\frac{π}{7})^4}$ Simplify
$$\frac{(\cos \frac{π}{7}-i\sin\frac{π}{7})^3}{(\cos\frac{π}{7}+i\sin\frac{π}{7})^4}$$
I used de Morvre's theorem to get to
$$\frac{(\cos \frac{3π}{7}-i\sin\frac{3π}{7})}{(\cos\frac{4π}{7}+i\sin\frac{4π}{7})}$$
How do you simplify from here?
The answer is meant to be $$\sin9\theta + i\cos\theta$$
| Hint:
$\cos\frac{3\pi}{7}=-\cos\frac{4\pi}{7}$, $\quad\sin\frac{3\pi}{7}=\sin\frac{4\pi}{7}.$
$$
\frac{(\cos \frac{3π}{7}-i\sin\frac{3π}{7})}{(\cos\frac{4π}{7}+i\sin\frac{4π}{7})}=\frac{(-\cos\frac{4π}{7}-i\sin\frac{4π}{7})}{(\cos\frac{4π}{7}+i\sin\frac{4π}{7})}=-1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if a,b,c,d are positive then $\frac{ab}{a+b}+\frac{cd}{c+d}\le \frac{(a+c)(b+d)}{a+b+c+d}$ Show that if a,b,c,d are positive then
$\frac{ab}{a+b}+\frac{cd}{c+d}\le \frac{(a+c)(b+d)}{a+b+c+d}$
I am stuck with this.
Thanks in advance!
| Hints: $\times (a+b)(c+d)(a+b+c+d)$ on the left side and right side.
| {
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"url": "https://math.stackexchange.com/questions/1248270",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Evaluating $ \int \frac{1}{5 + 3 \sin(x)} ~ \mathrm{d}{x} $.
What is the integral of: $\int \frac{1}{5+3\sin x}dx$
My attempt:
Using: $\tan \frac x 2=t$, $\sin x = \frac {2t}{1+t^2}$, $dx=\frac {2dt}{1+t^2}$ we have:
$\int \frac{1}{5+3\sin x}dx= 2\int \frac 1 {5t^2+6t+5}dt $
I'll expand the denominator: $5t^2+6t+5=5((t+\frac 3 5 )^2+1-\frac 1 4 \cdot (\frac 6 5)^2)=5((t+\frac 3 5)^2+0.64)$. So:
$2\int \frac 1 {5t^2+6t+5}dt = \frac 2 5 \int \frac 1{(t+\frac 3 5)^2+0.64}dt=\frac 2 5(\frac 5 4\arctan((t+\frac 3 5)\frac 5 4))=\frac 1 2 \arctan(\frac{5t+3}{4}) $
But if I'll place $\tan \frac x 2=t$ I won't be able to simplify it further and since the online calculator's answers don't have $\tan \frac x 2$ there, I believe I made a mistake. What is wrong with what I did and is there a better way to do it?
| Hint:
From what you got on that calculator, use the following identities to simplify it:
*
*$\sin(x)=2\sin(x/2)\cos(x/2)$
*$1+\cos(x)=2\cos^2(x/2)$
You'll see that the results are same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expected value with negative exponent I am trying to solve identify the expected value of a statistic that involves a fraction. I have simplified the expression to:
$E[\frac{1}{1+ \sum_i x_i}] = E[\frac{1}{1+ T}]$
However, I am not sure how to proceed. Is there anyway to simplify this through algebra, i.e. simplify the expression to the point that I have $E[x_i]$ and then substitute in a known expression for the expected value of $x_i$? Or, should I attempt to find the expected value of the expression by working with the pmf as
$\sum \frac{1}{1+ T} * f_T(t)$ ?
| For T distributed Negative Binomial (r,p) where $r>1$:
\begin{align*}
E \left( \frac{1}{1+T} \right) &= \sum \limits_{t=0}^{\infty} \frac{1}{1+t} \binom{r+t-1}{t} p^r (1-p)^t \\
&= \sum \limits_{t=0}^{\infty} \frac{1}{t+1} \cdot \frac{(r+t-1)!}{(t)!(r-1)!} p^r (1-p)^t \\
&= \sum \limits_{t=0}^{\infty} \frac{(r+t-1)!}{(t+1)!(r-1)!} p^r (1-p)^t \\
&= \sum \limits_{t=0}^{\infty} \frac{1}{r-1} \cdot \frac{(r+t-1)!}{(t+1)!(r-2)!} p^r (1-p)^t \\
&= \frac{1}{r-1} \sum \limits_{t=0}^{\infty} \binom{r+t-1}{t+1} p^r (1-p)^t \\
&= \frac{1}{r-1} \sum \limits_{t=0}^{\infty} \binom{r+(t+1)-2}{t+1} p^r (1-p)^t \\
&= \frac{1}{r-1} \sum \limits_{y=1}^{\infty} \binom{r+y-2}{y} p^r (1-p)^{y-1}, \,\,\, \text{where } y = t+1 \\
&= \frac{p}{(r-1)(1-p)} \sum \limits_{y=1}^{\infty} \binom{(r-1)+y-1}{y} p^{r-1} (1-p)^y \\
&= \frac{p}{(r-1)(1-p)} \left[ \sum \limits_{y=0}^{\infty} \binom{(r-1)+y-1}{y} p^{r-1} (1-p)^y \, - \, p^{r-1} \right] \\
&= \boxed{\frac{p ( 1 - p^{r-1} )}{(r-1)(1-p)}} \\
\end{align*}
Alternatively, if someone knows of a simple form for the Negative Binomial $n$th moment, then another solution might be found by the following Taylor expansion about 0:
$E \left( \frac{1}{1+T} \right) = E \left( 1 - T + T^2 - T^3 + \cdots \right) = \sum \limits_{n=0}^{\infty} (-1)^n E(T^n)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1254896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\int_{\sqrt{2}}^{\sqrt{5}} \frac{x^3}{\sqrt{x^2-1}} dx$ by substitution
$$\int_{\sqrt{2}}^{\sqrt{5}} \frac{x^3}{\sqrt{x^2-1}} dx$$
$u^2 = x^2 - 1$
I have worked out that $dx = du$ and that $u = x - 1$ so,
$\int\frac{x^3}{u} du$ - but I'm stuck at this stage.
Any help is appreciated, thanks!
| There is another step using integration by part
$$\int \frac{x^3}{\sqrt{x^2-1}}dx = \int x^2\frac{x}{\sqrt{x^2-1}}dx$$
You want to integrate $x(x^2-1)^{-1/2}$ (and differentiate $x^2$), consider product rule. (Substitution is using the idea of product rule, i.e. you can let $u=x^2-1$)
$$\int x(x^2-1)^{-1/2} = (x^2 - 1)^{1/2}dx$$
So the initial integration become
$$\int \frac{x^3}{\sqrt{x^2-1}}dx = x^2(x^2 - 1)^{1/2} - \int x(x^2-1)^{1/2}dx$$
Do integration by part again to yield final answer.
| {
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"url": "https://math.stackexchange.com/questions/1256083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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when index is irrational number with inequality
Let $x>0$, show that
$$x^{\sqrt{3}}+x^{\frac{\sqrt{3}}{2}}+1\ge 3\left(\dfrac{1+x}{2}\right)^{\sqrt{3}}$$
we consider
$$f(x)=2^{\sqrt{3}}(x^{\sqrt{3}}+x^{\dfrac{\sqrt{3}}{2}}+1)- 3(1+x)^{\sqrt{3}}$$
use computer in fact $f(x)\ge 0,x>0$,see Plot
we have
$$f'(x)=\dfrac{1}{2}\sqrt{3}\left(2^{\sqrt{3}}x^{\dfrac{\sqrt{3}}{2}-1}\left(
2x^{\dfrac{\sqrt{3}}{2}}+1\right)-6(x+1)^{\sqrt{3}-1}\right)$$
for this index irrational numbers,
what approaches do you think, I could take to solving the next step?
| first, we find a special point: $\dfrac{1}{3}> \left(\dfrac{1+x}{2}\right)^{\sqrt{3}} \implies x<2\left(\dfrac{1}{3}\right)^{\frac{1}{\sqrt{3}}}-1=0.0606 \implies x<0.06, $ the inequality is always true.
both side take log, we have: $\ln{\dfrac{x^{\sqrt{3}}+x^{\frac{\sqrt{3}}{2}}+1}{3}}\ge \ln{\left(\dfrac{1+x}{2}\right)^{\sqrt{3}}},f(x)=\ln{\dfrac{x^{\sqrt{3}}+x^{\frac{\sqrt{3}}{2}}+1}{3}}-\ln{\left(\dfrac{1+x}{2}\right)^{\sqrt{3}}}$
now we need to prove $f'(x)>0$ when $x>1$ and $f'(x)<0$ when $1>x>0.06$
$f'(x)=\dfrac{\sqrt{3}(2x^{\sqrt{3}}+x^{\frac{\sqrt{3}}{2}}(1-x)-2x)}{(x^{\sqrt{3}}(2x^2+2x)+x^{(\sqrt{3}/2)}(2x^2+2x)+2x^2+2x) }$
$g(x)=2x^{\sqrt{3}}+x^{\frac{\sqrt{3}}{2}}(1-x)-2x$
for $x>1$, $g(x)>0 \iff x^{\frac{\sqrt{3}}{2}}-\dfrac{x}{x^{\frac{\sqrt{3}}{2}}}>\dfrac{x-1}{2} \cap \dfrac{x}{x^{\frac{\sqrt{3}}{2}}}<x^{\frac{\sqrt{3}}{8}}\iff x^{\frac{\sqrt{3}}{2}}-x^{\frac{\sqrt{3}}{8}}>\dfrac{x-1}{2}$
for $0.06<x<1$,prove $g''>0$ which mean $g(x)$ will get max at $g(1)$ or $g(.06)$, if both end $\le 0 \implies g(x) \le 0$
I will not goto details because this is not so difficult.
Edit: a more straight way to prove $g(x)<0 \iff x^{\frac{\sqrt{3}}{2}}-\dfrac{x}{x^{\frac{\sqrt{3}}{2}}}<\dfrac{x-1}{2} $ is:
LHS$=h(x)=x^{\frac{\sqrt{3}}{2}}-\dfrac{x}{x^{\frac{\sqrt{3}}{2}}},h''(x)=x^{-2}(\dfrac{3}{2}-\sqrt{3})(x^{\frac{\sqrt{3}}{2}}-x^{1-\frac{\sqrt{3}}{2}})>0 \cap (x<1) \implies h(x) <y=\dfrac{h(1)-h(0.06)}{1-.06}(x-1)+h(1)=0.562(x-1)<0.5(x-1)$
the tricky here is when $x$ is small enough, $g(x)>0$, that is why we exclude small $x$ at beginning.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Trigonometric inequality in an obtuse triangle Let $ABC$ be an obtuse triangle with $A$ the obtuse angle. I conjecture that the following inequality is true $$\sin B + \sin C \le |\tan A|.$$
Show that it holds or give a counterexample.
| Since $B ,C , B+C < \frac{\pi}{2} $ we have $\frac{\cos B }{\cos (B +C )} >1$ and $\frac{\cos C }{\cos (B+C )} >1$ hence $$|\tan A|=\tan (B+C) =\frac{\cos B}{\cos (B+C)} \cdot \sin C + \frac{\cos C}{\cos (B+C)} \cdot \sin B >1\cdot \sin C +1\cdot \sin B =\sin C +\sin B .$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of integral solutions of $\text{xyz}=3000$ I want to find the number of integral solutions of the equation $$xyz=3000$$
I have been able to solve similar sums where the number on the right hand side was small enough to calculate all the factors of. Such as $xyz=24$ or $xyz=30$.
What is the proper method to solve such a problem when the number is too big to consider all the factors of it?
| Note that $3000 = 2^3 \cdot 3^1 \cdot 5^3$. Let $x=2^{x_1}3^{x_2}5^{x_3}$, $y=2^{y_1}3^{y_2}5^{y_3}$ and $z=2^{z_1}3^{z_2}5^{z_3}$. Then obtain the number of solutions to
$$x_1+x_2+x_3 = 3$$
$$y_1+y_2+y_3 = 1$$
$$z_1+z_2+z_3 = 3$$
The product of the number of non-negative solutions will give us the total number of solutions. Hence, in our case, the number of non-negative solutions to the first one is $\dbinom{3+3-1}{3-1} =10$, the second one is $\dbinom{1+3-1}{3-1} =3$ and the third one is $\dbinom{3+3-1}{3-1} =10$. Hence, total number of solutions is $10\cdot 3 \cdot 10 = 300$.
| {
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"url": "https://math.stackexchange.com/questions/1260093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Different Law of Cosines using Sine instead: $c^2 = a^2 + b^2 - 2ab\sqrt{1-\sin^2(\theta)}$ Playing around with Trig and the Law of Cosines (LoC), I came up with this formula given a triangle with sides $a$, $b$, $c$ where we are given $a$, $b$ and angle $\theta$ between them:
$$c^2 = a^2 + b^2 - 2ab\sqrt{1-\sin^2(\theta)}$$
Far from me the idea that I could've stumbled onto something no one's ever derived before, but I've never seen this formula and was just curious whether it has a name or is never considered because it offers no advantage over the LoC (needing the same amount of initial information) and is slightly more complicated.
Also, is my proof correct?
Here's my work; here I use $C$ for the angle:
$$c^2 = x^2 + h^2$$
$$h = a \sin(C)$$
$$h^2 = a^2 \sin^2(C)$$
$$x = b - (b-x)$$
$$(b-x) = \sqrt{a^2 - h^2} = \sqrt{a^2 - a^2 \sin^2(C)} = \sqrt{a^2 (1-\sin^2(C))}$$
$$x = b-a\sqrt{1-\sin^2(C)}$$
$$x^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C))$$
Therefore:
$$c^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C)) + a^2 \sin^2(C)$$
$$c^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C) + \sin^2(C))$$
$$c^2 = a^2 + b^2 - 2ab \sqrt{1-\sin^2(C)}$$
| Yours is same as the cosine rule. Recall that
$$c^2 = a^2+b^2-2ab\cos(C)$$
Now note that if $C$ is acute, we then have that $\cos(C) = \sqrt{1-\sin^2(C)}$. Hence, we obtain
$$c^2 = a^2+b^2-2ab\sqrt{1-\sin^2(C)}$$
Your proof is fine, though note that if $\angle{C}$ were to be obtuse, then writing $x$ as $b+(x-b)$ would be the right way to go about.
| {
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"url": "https://math.stackexchange.com/questions/1260795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Only finitely many $a, b$ such that $2+3^n+5^{n^2}=2^a7^b$ for some $n$? Let $(a,b)$ be a pair of positive integers such that $$2+3^n+5^{n^2}=2^a7^b$$ for some positive integer $n$. Is it true that there are only finitely many such pairs?
I don't know the answer to such question, but I guess that it is positive..
| No positive integer solutions exist to $2+3^n+5^{n^2}=2^a7^b$.
$n$ even $\,\Rightarrow\, 2+3^n+5^{n^2}\equiv 2+(-1)^{n^2}\equiv 2+1\equiv 0\not\equiv 2^a7^b\pmod{\!3}$.
$n$ odd $\,\Rightarrow\, 2+3^n+5^{n^2}\equiv 2+(-1)^n+1^{n^2}\equiv 2\equiv 2^a7^b\pmod{\!4}$, so $a=1$.
But $2+3^n+5^{n^2}\equiv 2+(-1)^{n^2}\equiv 1\not\equiv 2\cdot 7^b\equiv 2\cdot 1^b\equiv 2\pmod{\!3}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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An identity involving Bernoulli and Stirling numbers I was playing with some combinatorial sums and made an observation that I didn't know how to prove:
$$\forall n\in\mathbb N,\hspace{10px}\sum_{k=1}^n\frac{B_k\ S_1(n-1,\,k-1)}k=-\sum_{k=1}^n\frac{S_1(n,\,k)}{(k+1)\ n},$$
where $B_k$ are Bernoulli numbers and $S_1(n,\,k)$ are signed Stirling numbers of the first kind.
Could you please suggest any ideas how to prove it?
| Here is a re-write of my first proof with somewhat better aesthetics.
Suppose we seek to show that
$$\sum_{k=1}^n \frac{B_k}{k} (-1)^{n-k}
{n-1\brack k-1}
= -\frac{1}{n}
\sum_{k=1}^n \frac{1}{k+1} (-1)^{n-k} {n\brack k}.$$
Observe that using the EGF of the Stirling numbers of the first kind
we have
$${n\brack k}
= \frac{n!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\frac{1}{k!}\left(\log\frac{1}{1-z}\right)^k \; dz.$$
and using the EGF of the Bernoulli numbers we also have
$$B_k = \frac{k!}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{k+1}} \frac{w}{\exp(w)-1} \; dw.$$
The first of these controls the range and we may extend the sum to
infinity, obtaining for the LHS
$$ \frac{(n-1)! \times (-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n}
\int_{|w|=\epsilon} \frac{1}{\exp(w)-1}
\sum_{k\ge 1} (-1)^k \frac{1}{w^{k}}
\left(\log\frac{1}{1-z}\right)^{k-1}
\; dw \; dz.$$
This is
$$- \frac{(n-1)! \times (-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n}
\int_{|w|=\epsilon} \frac{1}{\exp(w)-1}
\frac{1}{w} \left(1+\frac{1}{w}
\log\frac{1}{1-z}\right)^{-1}
\; dw \; dz
\\ = - \frac{(n-1)! \times (-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n}
\int_{|w|=\epsilon} \frac{1}{\exp(w)-1}
\left(w+\log\frac{1}{1-z}\right)^{-1}
\; dw \; dz.$$
Extracting the residue from the pole at $w=0$ we obtain
$$- \frac{(n-1)! \times (-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n}
\left(\log\frac{1}{1-z}\right)^{-1} \; dz.$$
Next do the RHS, getting
$$-\frac{1}{n} (-1)^n\frac{n!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\sum_{k\ge 1} (-1)^k
\frac{1}{(k+1)!}\left(\log\frac{1}{1-z}\right)^k \; dz.$$
The inner term is
$$-\left(\log\frac{1}{1-z}\right)^{-1}
\sum_{k\ge 1} (-1)^{k+1}
\frac{1}{(k+1)!}\left(\log\frac{1}{1-z}\right)^{k+1}
\\ = -\left(\log\frac{1}{1-z}\right)^{-1}
\left(-1+ \log\frac{1}{1-z}
+ \exp\left(-\log\frac{1}{1-z}\right)\right)
\\ = -\left(\log\frac{1}{1-z}\right)^{-1}
\left(-1+ \log\frac{1}{1-z} + 1-z\right)
= -1 + z\left(\log\frac{1}{1-z}\right)^{-1}.$$
This gives for the RHS integral
$$ -\frac{1}{n} \frac{n!\times (-1)^n}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\left(-1 + z\left(\log\frac{1}{1-z}\right)^{-1}\right) \; dz.$$
When $n\ge 1$ this simplifies to
$$-\frac{(n-1)!\times (-1)^n}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
z\left(\log\frac{1}{1-z}\right)^{-1} \; dz$$
which is the same as the LHS, QED.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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